and Metric Topological Spaces
T. W. Krner
o
October 11, 2010
Small print The syllabus for the course is dened by the Faculty Board Schedules
(which are minimal for lecturing and maximal for examining). What is presented here
contains some results which it would not, in my opinion, be fair to set as book-work
although they could well appear as problems. In addition, I have included a small amount
of material which appears in other 1B courses. I should very much appreciate being told
of any corrections or possible improvements and might even part with a small reward to
A
the rst nder of particular errors. These notes are written in L TEX 2 and should be
available in tex, ps, pdf and dvi format from my home page
http://www.dpmms.cam.ac.uk/twk/
Contents
1 Preface
2
2 What is a metric?
3
3 Continuity and open sets for metric spaces
7
4 Closed sets for metric spaces
10
5 Topological spaces
12
6 More on topological structures
14
7 Hausdor spaces
18
8 Compactness
19
9 Products of compact spaces
24
10 Connectedness
25
1
11 Compactness in metric spaces
28
12 The language of neighbourhoods
30
13 Books
32
14 Exercises
33
15 Some hints
43
16 Some proofs
45
17 Executive summary
76
1
Preface
Within the last fty years the material in this course has been taught at
Cambridge in the fourth (postgraduate), third, second and rst years or left
to students to pick up for themselves. Under present arrangements students
may take the course either at the end of their rst year (before they have
met metric spaces in analysis) or at the end of their second year (after they
have met metric spaces).
Because of this, the rst third of the course presents a rapid overview
of metric spaces (either as revision or a rst glimpse) to set the scene for
the main topic of topological spaces. This arrangement is recognised in the
examination structure where the 12 lecture course is treated as though it
were an 8 lecture course.
The rst part of these notes states and discusses the main results of the
course. Usually, each statement is followed by directions to a proof in the
nal part of these notes. Whilst I do not expect the reader to nd all the
proofs by herself, I do ask that she tries to give a proof herself before looking
one up. Some of the more dicult theorems have been provided with hints
as well as proofs.
In my opinion, the two sections on compactness are the deepest part of
the course and the reader who has mastered the proofs of the results therein
is well on the way to mastering the whole course.
May I repeat that, as I said in the small print, I welcome corrections and
comments.
2
2
What is a metric?
If I wish to travel from Cambridge to Edinburgh, then I may be interested
in one or more of the following numbers.
(1) The distance, in kilometres, from Cambridge to Edinburgh as the
crow ies.
(2) The distance, in kilometres, from Cambridge to Edinburgh by road.
(3) The time, in minutes, of the shortest journey from Cambridge to
Edinburgh by rail.
(4) The cost, in pounds, of the cheapest journey from Cambridge to
Edinburgh by rail.
Each of these numbers is of interest to someone and none of them is easily
obtained from another. However, they do have certain properties in common
which we try to isolate in the following denition.
Denition 2.1. Let X be a set1 and d : X 2 R a function with the following
properties:(i) d(x, y ) 0 for all x, y X .
(ii) d(x, y ) = 0 if and only if x = y .
(iii) d(x, y ) = d(y, x) for all x, y X .
(iv) d(x, y ) + d(y, z ) d(x, z ) for all x, y, z X . (This is called the
triangle inequality after the result in Euclidean geometry that the sum of
the lengths of two sides of a triangle is at least as great as the length of the
third side.)
Then we say that d is a metric on X and that (X, d) is a metric space.
You should imagine the author muttering under his breath
(i) Distances are always positive.
(ii) Two points are zero distance apart if and only if they are the same
point.
(iii) The distance from A to B is the same as the distance from B to A.
(iv) The distance from A to B via C is at least as great as the distance
from A to B directly.
Exercise 2.2. If d : X 2 R is a function with the following properties:1
We thus allow X = . This is purely a question of taste. If we did not allow this
possibility, then, every time we dened a metric space (X, d), we would need to prove
that X was non-empty. If we do allow this possibility, and we prefer to reason about
non-empty spaces, then we can begin our proof with the words If X is empty, then the
result is vacuously true, so we may assume that X is non-empty. (Of course, the result
may be false for X = , in which case the statement of the theorem must include the
condition X = .)
3
(ii) d(x, y ) = 0 if and only if x = y ,
(iii) d(x, y ) = d(y, x) for all x, y X ,
(iv) d(x, y ) + d(y, z ) d(x, z ) for all x, y, z X ,
show that d is a metric on X . [Thus condition (i) of the denition is redundant.]
Proof. See page 45.
Exercise 2.3. Let X be the set of towns on the British railway system. Consider the d corresponding to the examples (1) to (4) and discuss informally
whether conditions (i) to (iv) apply. [An open ended question like this will
be more useful if tackled in a spirit of good will.]
Exercise 2.4. Let X = {a, b, c} with a, b and c distinct. Write down
functions dj : X 2 R satisfying condition (i) of Denition 2.1 such that:(1) d1 satises conditions (ii) and (iii) but not (iv).
(2) d2 satises conditions (iii) and (iv) and d2 (x, y ) = 0 implies x = y ,
but it is not true that x = y implies d2 (x, y ) = 0.
(3) d3 satises conditions (iii) and (iv) and x = y implies d3 (x, y ) = 0.
but it is not true that d3 (x, y ) = 0 implies x = y .
(4) d4 satises conditions (ii) and (iv) but not (iii).
You should verify your statements.
Solution. See page 45.
We give another axiom grubbing exercise as Exercise 14.1.
Exercise 2.5. Let X be a set and : X 2 R a function with the following
properties.
(i) (x, y ) 0 for all x, y X .
(ii) (x, y ) = 0 if and only if x = y .
(iv) (x, y ) + (y, z ) (x, z ) for all x, y, z X .
Show that, if we set d(x, y ) = (x, y ) + (y, x), then (X, d) is a metric space.
Here are some examples of metric spaces. You have met (or you will
meet) the concept of a normed vector space both in algebra and analysis
courses.
Denition 2.6. Let V be a vector space over F (with F = R or F = C) and
N : V R a map such that, writing N (u) = u , the following results hold.
(i) u 0 for all u V .
(ii) If u = 0, then u = 0.
(iii) If F and u V , then u = || u .
(iv) [Triangle law.] If u, v V , then u + v u + v .
Then we call
a norm and say that (V,
) is a normed vector space.
4
Exercise 2.7. By putting = 0 in Denition 2.6 (iii), show that 0 = 0.
Any normed vector space can be made into a metric space in a natural
way.
Lemma 2.8. If (V,
) is a normed vector space, then the condition
d(u, v) = u v
denes a metric d on V .
Proof. The easy proof is given on page 46.
Many (but not all) metrics that we meet in analysis arise in this way.
However, not all metrics can be derived from norms in this way. Here is
a metric that turns out to be more important and less peculiar than it looks
at rst sight.
Denition 2.9. If X is a set and we dene d : X 2 R by
d(x, y ) =
0 if x = y,
1 if x = y,
then d is called the discrete metric on X .
Lemma 2.10. The discrete metric on X is indeed a metric.
Proof. The easy proof is given on page 47.
The next result, although easy, is very denitely not part of the course,
so I leave it as an exercise for the interested reader.
Exercise 2.11. (i) If V is a vector space over R and d is a metric derived from a norm in the manner described above, then, if u V we have
d(0, 2u) = 2d(0, u).
(ii) If V is non-trivial (i.e. not zero-dimensional) vector space over R
and d is the discrete metric on V , then d cannot be derived from a norm on
V.
In algebra you have learnt (or you will learn) about inner product spaces.
You have learnt (or you will learn) that every inner product gives rise to a
norm in a natural way. Most norms in analysis do not arise in this way2 but
the few that do are very important.
2
This is not part of the course, but see Exercise 14.2 if you are interested.
5
Denition 2.12. If x Rn , we write
1/2
n
x
2
x2
j
=
,
j =1
where the positive square root is taken. We call
Rn .
2
the Euclidean norm on
The reader should prove the next lemma before proceeding further. (Pay
particular attention to the triangle inequality. In my opinion, the easiest
proof uses inner products but this is only an opinion and you may ignore it.).
Lemma 2.13. The Euclidean norm on Rn is indeed a norm.
The metric derived from the Euclidean norm is called the Euclidean metric. You should test any putative theorems on metric spaces on both Rn with
the Euclidean metric and Rn with the discrete metric.
Exercise 2.14. [The counting metric.] If E is a nite set and E is the
collection of subsets of E , we write card C for the number of elements in C
and
d(A, B ) = card AB.
Show that d is a metric on E . The reader may be inclined to dismiss this metric as uninteresting but it plays an important role (as the Hamming metric)
in the Part II course Codes and Cryptography.
Here are two metrics which are included simply to show that metrics do
not always look as simple as the ones above. I shall use them as examples
once or twice but they do not form part of standard mathematical knowledge
and you do not have to learn their denition.
Denition 2.15. (i) If we dene d : R2 R2 R by
u
d(u, v) =
2
+ v 2,
0
if u = v,
if u = v,
then d is called the British Rail express metric. [To get from A to B travel
via London.]
(ii) If we dene d : R2 R2 R by
d(u, v) =
uv 2
u 2+ v
2
if u and v are linearly dependent,
otherwise,
then d is called the British Rail stopping metric. [To get from A to B travel
via London unless A and B are on the same London route.]
6
(Recall that u and v are linearly dependent if u = v for some real
and/or v = 0.)
Exercise 2.16. Show that the British Rail express metric and the British
Rail stopping metric are indeed metrics.
Solution. On page 47 we show that the British Rail stopping metric is indeed
a metric. The British Rail express metric can be dealt with similarly.
In the long Exercise 14.26 we look at a metric which plays an important
role in complex analysis and in geometry.
3
Continuity and open sets for metric spaces
Some denitions and results transfer essentially unchanged from classical
analysis on R to metric spaces. Recall the classical denition of continuity.
Denition 3.1. [Old denition.] A function f : R R is called continuous if, given t R and > 0, we can nd a (t, ) > 0 such that
|f (t) f (s)| < whenever |t s| < (t, ).
It is not hard to extend this denition to our new, wider context.
Denition 3.2. [New denition.] Let (X, d) and (Y, ) be metric spaces.
A function f : X Y is called continuous if, given t X and > 0, we can
nd a (t, ) > 0 such that
(f (t), f (s)) < whenever d(t, s) < (t, ).
It may help you grasp this denition if you read (f (t), f (s)) as the
distance from f (t) to f (s) in Y and d(t, s) as the distance from t to s in
X .
Lemma 3.3. [The composition law.] If (X, d) and (Y, ) and (Z, ) are
metric spaces and g : X Y , f : Y Z are continuous, then so is the
composition f g .
Proof. This is identical to the one we met in classical analysis. If needed,
details are given on page 48.
Exercise 3.4. Let R and R2 have their usual (Euclidean) metric.
(i) Suppose that f : R R and g : R R are continuous. Show that the
map (f, g ) : R2 R2 is continuous.
(ii) Show that the map M : R2 R given by M (x, y ) = xy is continuous.
(iii) Use the composition law to show that the map m : R2 R given by
m(x, y ) = f (x)g (y ) is continuous.
7
Solution. See page 48.
Exercise 3.4 may look perverse at rst sight but, in fact, we usually show
functions to be continuous by considering them as compositions of simpler
functions rather than using the denition directly. Think about
x log 2 + sin
1
1 + x2
.
If you are interested, we continue the chain of thought in Exercise 14.3. If
you are not interested or are mildly confused by all this, just ignore this
paragraph.
Just as there are well behaved and badly behaved functions between
spaces so there are well behaved and badly behaved subsets of spaces. In
classical analysis and analysis on metric spaces the notion of continuous function is suciently wide to give us a large collection of interesting functions
and suciently narrow to ensure reasonable behaviour3 . In introductory
analysis we work on R with the Euclidean metric and only consider subsets
in the form of intervals. Once we move to R2 with the Euclidean metric, it
becomes clear that there is no appropriate analogue to intervals. (We want
some rectangles to be well behaved but we also want to talk about discs and
triangles and blobs.)
Cantor identied two particular classes of well behaved sets. We start
with open sets.
Denition 3.5. Let (X, d) be a metric space. We say that a subset E is
open in X if, whenever e E , we can nd a > 0 (depending on e) such
that
x E whenever d(x, e) < .
Suppose we work in R2 with the Euclidean metric. If E is an open set
then any point e in E is the centre of a disc of strictly positive radius all of
whose points lie in E . If we are suciently short sighted, every point that we
can see from e lies in E . This property turns out to be a key to many proofs
in classical analysis (remember that in the proof of Rolles theorem it was
vital that the maximum did not lie at an end point) and complex analysis
(where we examine functions analytic on an open set ).
Here are a couple of simple examples of an open set and a simple example
of a set which is not open.
3
Sentences like this are not mathematical statements but many mathematicians nd
them useful.
8
Example 3.6. (i) Let (X, d) be a metric space. If r > 0, then
B (x, r) = {y : d(x, y ) < r}
is open.
(ii) If we work in Rn with the Euclidean metric, then the one point set
{x} is not open.
(iii) If (X, d) is a discrete metric space, then
{x} = B (x, 1/2)
and all subsets of X are open.
Proof. See page 49.
We call B (x, r) the open ball with centre x and radius r. The following
result is very important for the course but is also very easy to check.
Theorem 3.7. If (X, d) is a metric space then the following statements are
true.
(i) The empty set and the space X are open.
(ii) If U is open for all A, then A U is open. (In other words
the union of open sets is open.)
(iii) If Uj is open for all 1 j n, then n=1 Uj is open.
j
Proof. See page 49.
It is important to realise that we place no restriction on the size of A
in (ii). In particular, A could be uncountable. However, conclusion (iii)
cannot be extended.
Example 3.8. Let us work in Rn with the usual metric. Then B (x, 1/j ) is
open but B (x, 1/j ) = {x} is not.
j =1
Proof. See Example 3.6.
There is a remarkable connection between the notion of open sets and
continuity.
Theorem 3.9. Let (X, d) and (Y, ) be metric spaces. A function f : X Y
is continuous if and only if f 1 (U ) is open in X whenever U is open in Y .
Proof. See page 50.
Note that the theorem does not work in the opposite direction.
9
Example 3.10. Let X = R and d be the discrete metric. Let Y = R and
be the usual (Euclidean) metric.
(i) If we dene f : X Y by f (x) = x, then f is continuous but there
exist open sets U in X such that f (U ) is not open.
(ii)If we dene g : Y X by g (y ) = y , then g is not continuous but g (V )
is open in X whenever V is open in Y .
Proof. Very easy, but see page 51 if you need.
The message of this example is reinforced by the more complicated Exercise 14.4.
Observe that Theorem 3.9 gives a very neat proof of the composition law.
Theorem 3.3. If (X, d) and (Y, ) and (Z, ) are metric spaces and g : X
Y , f : Y Z are continuous, then so is the composition f g .
New proof. If U is open in Z , then, by continuity, f 1 (U ) is open in Y
and so, by continuity, (f g )1 (U ) = g 1 f 1 (U ) is open in X . Thus f g is
continuous.
This conrms our feeling that the ideas of this chapter are on the right
track.
We nish with an exercise which may be omitted at rst reading but
which should be done at some time as examples of what open sets can look
like.
Exercise 3.11. Consider R2 . For each of the British rail express and British
rail stopping metrics:(i) Describe the open balls. (Consider both large and small radii.)
(ii) Describe the open sets as well as you can. (There is a nice description
for the British rail express metric.) Give reasons for your answers.
Solution. See page 51.
4
Closed sets for metric spaces
The second class of well behaved sets identied by Cantor were the closed
sets. In order to dene closed sets in metric spaces we need a notion of limit.
Fortunately, the classical denition generalises without diculty.
Denition 4.1. Consider a sequence xn in a metric space (X, d). If x X
and, given > 0, we can nd an integer N 1 (depending on ) such that
d(xn , x) < for all n N ,
then we say that xn x as n .
10
Lemma 4.2. Consider a metric space (X, d). If a sequence xn has a limit,
then that limit is unique.
Proof. The simple proof is given on page 52.
Denition 4.3. Let (X, d) be a metric space. A set F in X is said to be
closed if, whenever xn F and xn x as n , it follows that x F .
The following exercise is easy but instructive.
Exercise 4.4. (i) If (X, d) is any metric space, then X and are both open
and closed.
(ii) If we consider R with the usual metric and take b > a, then [a, b] is
closed but not open, (a, b) is open but not closed and [a, b) is neither open
nor closed.
(iii) If (X, d) is a metric space with discrete metric d, then all subsets of
X are both open and closed.
It is easy to see why closed sets will be useful in those parts of analysis
which involve taking limits. The reader will recall theorems in elementary
analysis (for example the boundedness of continuous functions) which were
true for closed intervals but not for other types of intervals.
Life is made much easier by the very close link between the notions of
closed and open sets given by our next theorem.
Theorem 4.5. Let (X, d) be a metric space. A set F in X is closed if and
only if its complement is open.
Proof. We give the proof on page 53.
We can now deduce properties of closed sets from properties of open sets
by complementation. In particular, we have the following complementary
versions of Theorems 3.7 and 3.9
Theorem 4.6. If (X, d) is a metric space, then the following statements are
true.
(i) The empty set and the space X are closed.
(ii) If F is closed for all A, then A F is closed. (In other words
the intersection of closed sets is closed.)
(iii) If Fj is closed for all 1 j n, then n=1 Fj is closed.
j
Proof. See page 53.
Theorem 4.7. Let (X, d) and (Y, ) be metric spaces. A function f : X Y
is continuous if and only if f 1 (F ) is closed in X whenever F is closed in
Y.
Proof. See page 54.
11
5
Topological spaces
We now investigate general objects which have the structure described by
Theorem 3.7.
Denition 5.1. Let X be a set and a collection of subsets of X with the
following properties.
(i) The empty set and the space X .
(ii) If U for all A, then A U .
(iii) If Uj for all 1 j n, then n=1 Uj .
j
Then we say that is a topology on X and that (X, ) is a topological
space.
Theorem 5.2. If (X, d) is a metric space, then the collection of open sets
forms a topology.
Proof. This is Theorem 3.7.
If (X, d) is a metric space we call the collection of open sets the topology
induced by the metric.
If (X, ) is a topological space we extend the notion of open set by calling
the members of open sets. The discussion above ensures what computer
scientists call downward compatibility.
Exercise 5.3. If (X, d) is a metric space with the discrete metric, show that
the induced topology consists of all the subsets of X .
We call the topology consisting of all subsets of X the discrete topology
on X .
Exercise 5.4. If X is a set and = {, X }, then is a topology.
We call {, X } the indiscrete topology on X .
Exercise 5.5. (i) If F is a nite set and (F, d) is a metric space show that
the induced topology is the discrete topology.
(ii) If F is a nite set with more than one point, show that the indiscrete
topology is not induced by any metric.
You should test any putative theorems on topological spaces on the discrete topology and the indiscrete topology, Rn with the topology derived from
the Euclidean metric and [0, 1] with the topology derived from the Euclidean
metric.
The following exercise is tedious but instructive (the tediousness is the
instruction).
12
Exercise 5.6. Write P (Y ) for the collection of subsets of Y . If X has three
elements, how many elements does P P (X ) have?
How many topologies are there on X ?
Solution. See page 54.
The idea of downward compatibility suggests turning Theorem 3.9 in a
denition.
Denition 5.7. Let (X, ) and (Y, ) be topological spaces. A function f :
X Y is said to be continuous if and only if f 1 (U ) is open in X whenever
U is open in Y .
Theorem 3.9 tells us that if (X, d) and (Y, ) are metric spaces the notion
of a continuous function f : X Y is the same whether we consider the
metrics or the topologies derived from them.
The proof of Theorem 3.3 given on page 10 carries over unchanged to give
the following generalisation.
Theorem 5.8. If (X, ) and (Y, ) and (Z, ) are topological spaces and
g : X Y , f : Y Z are continuous, then so is the composition f g .
Downward compatibility suggests the denition of a closed set for a topological space based on Theorem 4.5.
Denition 5.9. Let (X, ) be a topological space. A set F in X is said to
be closed if its complement is open.
Theorem 4.5 tells us that if (X, d) is a metric space the notion of a closed
set is the same whether we consider the metric or the topology derived from
it.
Just as in the metric case, we can deduce properties of closed sets from
properties of open sets by complementation.In particular, the same proofs
as we gave in the metric case give the following extensions of Theorems 4.6
and 4.7
Theorem 5.10. If (X, ) is a topological space, then the following statements
are true.
(i) The empty set and the space X are closed.
(ii) If F is closed for all A, then A F is closed. (In other words,
the intersection of closed sets is closed.)
(iii) If Fj is closed for all 1 j n, then n=1 Fj is closed.
j
Theorem 5.11. Let (X, ) and (Y, ) be topological spaces. A function f :
X Y is continuous if and only if f 1 (F ) is closed in X whenever F is
closed in Y .
13
6
More on topological structures
Two groups are the same for the purposes of group theory if they are (group)
isomorphic. Two vector spaces are the same for the purposes of linear algebra
if they are (vector space) isomorphic. When are two topological spaces (X, )
and (Y, ) the same for the purposes of topology? In other words, when does
there exist a bijection between X and Y in which open sets correspond to
open sets, and the grammar of topology (things like union and inclusion)
is preserved? A little reection shows that the next denition provides the
answer we want. (Exercise 14.8 is vaguely relevant.)
Denition 6.1. We say that two topological spaces (X, ) and (Y, ) are
homeomorphic if there exists a bijection : X Y such that and 1 are
continuous. We call a homeomorphism.
The following exercise acts as useful revision of concepts learnt last year.
Exercise 6.2. Show that homeomorphism is an equivalence relation on topological spaces.
Homeomorphism only implies equivalence for the purposes of topology. To
emphasise this, we introduce a couple of related ideas which are fundamental
to analysis on metric spaces but which will only be referred to here in this
course.
Denition 6.3. (i) If (X, d) is a metric space, we say that a sequence xn in
X is Cauchy if, given > 0, we can nd an N0 () with
d(xn , xm ) < whenever n, m N0 ().
(ii) We say that a metric space (X, d) is complete if every Cauchy sequence converges.
Example 6.4. Let X = R and let d be the usual metric on R. Let Y = (0, 1)
(the open interval with end points 0 and 1) and let be the usual metric on
(0, 1). Then (X, d) and (Y, ) are homeomorphic as topological spaces but
(X, d) is complete and (Y, ) is not.
Proof. See page 55.
We say that completeness is not a topological property.
In group theory we usually prove that two groups are isomorphic by
constructing an explicit isomorphism and that two groups are not isomorphic
by nding a group property exhibited by one but not by the other. Similarly
14
in topology we usually prove that two topological spaces are homeomorphic
by constructing an explicit homeomorphism and that two topological spaces
are not homeomorphic by nding a topological property exhibited by one
but not by the other. Later in this course we will meet some topological
properties like being Hausdor and compactness and you will be able to
tackle Exercise 14.20.
We also want to be able to construct new topological spaces from old. To
do this we we make use of a simple but useful lemma.
Lemma 6.5. Let X be a space and let H be a collection of subsets of X .
Then there exists a unique topology H such that
(i) H H, and
(ii) if is a topology with H, then H .
Proof. The proof, which follows the standard pattern for such things, is given
on page 55.
We call H the smallest (or coarsest) topology containing H.
Lemma 6.6. Suppose that A is non-empty and the spaces (X , ) are topological spaces and we have maps f : X X [ A]. Then there is a
smallest topology on X for which the maps f are continuous.
Proof. A topology on X makes all the f continuous if and only if it
contains
H = {f 1 (U ) : U , A}.
Now apply Lemma 6.5.
Recall that, if Y X , then the inclusion map j : Y X is dened by
j (y ) = y for all y Y .
Denition 6.7. If (X, ) is a topological space and Y X then the subspace
topology Y on Y induced by is the smallest topology on Y for which the
inclusion map is continuous.
Lemma 6.8. If (X, ) is a topological space and Y X , then the subspace
topology Y on Y is the collection of sets Y U with U .
Proof. The very easy proof is given on page 56.
Exercise 6.9. (i) If (X, ) is a topological space and Y X is open, show
that the subspace topology Y on Y is the collection of sets U with U Y .
(ii) Consider R with the usual topology (that is, the topology derived
from the Euclidean metric). If Y = [0, 1], show that [0, 1/2) Y but
[0, 1/2) .
/
15
Exercise 6.10. Let (X, d) be a metric space, Y a subset of X and dY the
metric d restricted to Y (formally, dY : Y 2 R is given by dY (x, y ) = d(x, y )
for x, y Y ). Then if we give X the topology induced by d, the subspace
topology on Y is identical with the topology induced by dY .
[This is an exercise in stating the obvious.]
Next recall that if X and Y are sets the projection maps X : X Y X
and Y : X Y Y are given by
X (x, y ) = x,
Y (x, y ) = y.
Denition 6.11. If (X, ) and (Y, ) are topological spaces, then the product
topology on X Y is the smallest topology on X Y for which the projection
maps X and Y are continuous.
Lemma 6.12. Let (X, ) and (Y, ) be topological spaces and the product
topology on X Y . Then O if and only if, given (x, y ) O, we can nd
U and V such that
(x, y ) U V O.
Proof. See page 56.
The next remark is useful for proving results like those in Exercise 6.14.
Lemma 6.13. Let 1 and 2 be two topologies on the same space X .
(i) We have 1 2 if and only if, given x U 1 , we can nd V 2
such that x V U .
(ii)We have 1 = 2 if and only if, given x U 1 , we can nd V 2
such that x V U and given x U 2 we can nd V 1 such that
x V U.
Proof. The easy proof is given on Page 57
Exercise 6.14. Let (X1 , d1 ) and (X2 , d2 ) be metric spaces. Let be the
product topology on X1 X2 where Xj is given the topology induced by dj
[j = 1, 2].
Dene k : (X1 X2 )2 R by
1 ((x, y ), (u, v )) = d1 (x, u),
2 ((x, y ), (u, v )) = d1 (x, u) + d2 (y, v ),
3 ((x, y ), (u, v )) = max(d1 (x, u), d2 (y, v )),
4 ((x, y ), (u, v )) = (d1 (x, u)2 + d2 (y, v )2 )1/2 .
Establish that 1 is not a metric and that 2 , 3 and 4 are. Show that
each of the j with 2 j 4 induce the product topology on X1 X2 .
16
It is easy to extend our denitions and results to any nite product of
topological spaces. In fact, it is not dicult to extend our denition to the
product of an innite collection of topological spaces but I feel that it is
important for the reader to concentrate on rst thoroughly understanding
the nite product case and I have relegated the innite case to an exercise
(Exercise 14.10).
We conclude this chapter by looking briey at the quotient topology. This
will not play a major part in our course and the reader should not worry too
much about it.
If is an equivalence relation on a set X then we know from previous
courses that it gives rise to equivalence classes
[x] = {y X : y x}.
There is a natural map q from X to the space X/ of equivalence classes
given by q (x) = [x]. When we dened the subspace and product topologies
we used natural maps from the new spaces to the old spaces. Here we have
a natural map from the old space to the new, so our denition has to take a
dierent form.
Since intersection and union behave well under inverse mappings it is easy
to check the following statement.
Lemma 6.15. Let (X, ) be a topological space and an equivalence relation
on X . Write q for the map from X to the quotient space X/ given by
q (x) = [x]. Then
= {U X/ : q 1 (U ) }
is a topology.
Denition 6.16. Under the assumptions and with the notation of Lemma 6.15
we call the quotient topology on X/ .
The following is just a restatement of the denition.
Lemma 6.17. Under the assumptions and with the notation of Lemma 6.15,
the quotient topology consists of the sets U such that
[x] .
[x]U
Later we shall give an example (Exercise 9.7) of a nice quotient topology.
Exercise 14.24, which requires ideas from later in the course, is an example
of really nasty quotient topology.
17
In general, the quotient topology can be extremely unpleasant (basically
because equivalence relations form a very wide class) and although nice equivalence relations sometimes give useful quotient topologies you should always
think before using one. Exercises 14.11 and 14.12 give some further information.
7
Hausdor spaces
When we work in a metric space we make repeated use of the fact that if
d(x, y ) = 0 then x = y . The metric is powerful enough to separate points.
The indiscrete topology, on the other hand, clearly can not separate points.
When Hausdor rst crystallised the modern idea of a topological space
he included an extra condition to ensure separation of points. It was later
discovered that topologies without this extra condition could be useful so it
is now considered separately.
Denition 7.1. A topological space (X, ) is called Hausdor if, whenever
x, y X and x = y , we can nd U, V such that x U , y V and
U V = .
In the English educational system it is traditional to draw U and V as
little huts containing x and y and to say that x and y are housed o from
each other.
The next exercise requires a one line answer but you should write that
line down.
Exercise 7.2. Show that, if (X, d) is a metric space, then the derived topology
is Hausdor.
Although we defer the discussion of neighbourhoods in general to towards
the end of the course, it is natural to introduce the following locution here.
Denition 7.3. If (X, ) is a topological space and x U , we call U an
open neighbourhood of x.
Exercise 7.4. If (X, ) is a topological space, then a subset A of X is open
if and only if every point of A has an open neighbourhood U A.
Proof. The easy proof is given on page 58.
Lemma 7.5. If (X, ) is a Hausdor space, then the one point sets {x} are
closed.
Proof. The easy proof is given on page 58.
18
The following exercise shows that the converse to Lemma 7.5 is false and
that, if we are to acquire any intuition about topological spaces, we will need
to study a wide range of examples.
Exercise 7.6. Let X be innite (we could take X = Z or X = R). We say
that a subset E of X lies in if either E = or X \ E is nite. Show that
is a topology and that every one point set {x} is closed but that (X, ) is
not Hausdor.
What happens if X is nite?
Proof. See page 58.
It is easy to give examples of topologies which are not derived from metrics. It is somewhat harder to give examples of Hausdor topologies which
are not derived from metrics. An example is given in Exercise 14.13.
The next two lemmas are very useful.
Lemma 7.7. If (X, ) is a Hausdor topological space and Y X , then Y
with the subspace topology is also Hausdor.
Proof. The easy proof is given on page 59.
Lemma 7.8. If (X, ) and (Y, ) are Hausdor topological spaces, then X Y
with the product topology is also Hausdor.
Proof. The proof is easy (but there is one place where you can make a silly
mistake). We give it on page 59.
Exercise 14.12 shows that, even when the original topology is Hausdor,
the resulting quotient topology need not be.
8
Compactness
Halmos says somewhere that if an idea is used once it is a trick, if used twice
it is a method, if used three times a theorem but if used four times it becomes
an axiom.
Several important theorems in analysis hold for closed bounded intervals. Heine used a particular idea to prove one of these. Borel isolated the
idea as a theorem (the HeineBorel theorem), essentially Theorem 8.5 below. Many treatments of analysis (for example, Hardys Pure Mathematics )
use the HeineBorel theorem as a basic tool. The notion of compactness
represents the last stage in in the Halmos progression.
19
Denition 8.1. A topological space (X, ) is called compact if, whenever we
have a collection U of open sets [ A] with A U = X , we can nd a
nite subcollection U(1) , U(2) , . . . , U(n) with (j ) A [1 j n] such
that n=1 U(j ) = X .
j
Denition 8.2. If (X, ) is a topological space, then a subset Y is called
compact if the subspace topology on Y is compact.
The reader should have no diculty in combining these two denitions
to come up with the following restatement,
Lemma 8.3. If (X, ) is a topological space, then a subset Y is compact if,
whenever we have a collection U of open sets [ A] with U Y , we
can nd a nite subcollection U(1) , U(2) , . . . , U(n) with (j ) A [1 j
n] such that n=1 U(j ) Y .
j
In other words, a set is compact if any cover by open sets has a nite
subcover.
The reader is warned that compactness is a subtle property which requires
time and energy to master4 . (At the simplest level, a substantial minority
of examinees fail to get the denition correct.) Up to this point most of the
proofs in this course have been simple deductions from denitions. Several
of our theorems on compactness go much deeper and have quite intricate
proofs.
Here are some simple examples of compactness and non-compactness.
Exercise 8.4. (i) Show that, if X is nite, every topology on X is compact.
(ii) Show that the discrete topology on a set X is compact if and only if
X is nite.
(iii) Show that the indiscrete topology is always compact.
(iv) Show that the topology described in Exercise 7.6 is compact.
(v) Let X be uncountable (we could take X = R). We say that a subset
A of X lies in if either A = or X \ A is countable. Show that is a
topology but that (X, ) is not compact.
Solution. We give a partial solution for parts (iv) and (v) on page 59.
We now come to our rst major theorem.
Theorem 8.5. [The HeineBorel Theorem.] Let R be given its usual
topology (that is to say the topology derived from the usual Euclidean metric).
Then the closed bounded interval [a, b] is compact.
4
My generation only reached compactness after a long exposure to the classical Heine
Borel theorem.
20
Proof. I give a hint on page 43 and a proof on 60. An alternative proof,
which is much less instructive, is given on page 29.
Lemma 8.3 gives the following equivalent statement.
Theorem 8.6. Let [a, b] be given its usual topology (that is to say the topology derived from the usual Euclidean metric). Then the derived topology is
compact.
We now have a couple of very useful results.
Theorem 8.7. A closed subset of a compact set is compact. [More precisely,
if E is compact and F closed in a given topology, then, if F E , it follows
that F is compact.]
Proof. This is easy if you look at it the right way. See page 61.
Theorem 8.8. If (X, ) is Hausdor, then every compact set is closed.
Proof. This is harder, though it becomes easier if you realise that you must
use the fact that is Hausdor (see Example 8.9 below). We give a hint on
page 43 and a proof on page 61.
Example 8.9. Give an example of a topological space (X, ) and a compact
set in X which is not closed.
Proof. There is a topological space with two points which will do. See
page 62.
Combining the HeineBorel theorem with Theorems 8.7 and 8.8 and a
little thought, we get a complete characterisation of the compact subsets of
R (with the standard topology).
Theorem 8.10. Consider (R, ) with the standard (Euclidean) topology. A
set E is compact if and only if it is closed and bounded (that is to say, there
exists a M such that |x| M for all x E ).
Proof. The easy proof is given on page 62.
In Example 3.10 we saw that the continuous image of an open set need
not be open. It also easy to see that the continuous image of a closed set
need not be closed.
Exercise 8.11. Let R have the usual metric. Give an example of a continuous injective function f : R R such that f (R) is not closed.
Hint. Look at the solution of Example 6.4 if you need a hint.
21
However, the continuous image of a compact set is always compact.
Theorem 8.12. Let (X, ) and (Y, ) be topological spaces and f : X Y a
continuous function. If K is a compact subset of X , then f (K ) is a compact
subset of Y .
Proof. This is easy if you look at it the right way. See page 62.
This result has many delightful consequences. Recall, for example, that
the quotient topology X/ is dened in such a way that the quotient map
q : X X/ is continuous. Since q (X ) = X/ , Theorem 8.12 gives us a
positive property of the quotient topology.
Theorem 8.13. Let (X, ) be a compact topological space and an equivalence relation on X . Then the quotient topology on X/ is compact.
The next result follows at once from our characterisation of compact sets
for the real line with the usual topology.
Theorem 8.14. Let R have the usual metric. If K is a closed and bounded
subset of R and f : K R is continuous, then f (K ) is closed and bounded.
This gives a striking extension of one of the crowning glories of a rst
course in analysis.
Theorem 8.15. Let R have the usual metric. If K is a closed and bounded
subset of R and f : K R is continuous, then f is bounded and attains its
bounds.
Proof. The straightforward proof is given on page 63.
Theorem 8.15 is complemented by the following observation.
Exercise 8.16. Let R have the usual metric. If K is subset of R with the
property that, whenever f : K R is continuous, f is bounded and attains
its bounds, it follows that K is closed and bounded.
Proof. See page 63.
Theorem 8.15 has the following straightforward generalisation whose proof
is left to the reader.
Theorem 8.17. If K is a compact space and f : K R is continuous then
f is bounded and attains its bounds.
We also have the following useful result.
22
Theorem 8.18. Let (X, ) be a compact and (Y, ) a Hausdor topological
space. If f : X Y is a continuous bijection, then it is a homeomorphism.
Proof. There is a hint on page 43 and a proof on page 63.
Theorem 8.18 is illuminated by the following almost trivial remark.
Lemma 8.19. Let 1 and 2 be topologies on the same space X . The identity
map
: (X, 1 ) (X, 2 )
from X with topology 1 to X with topology 2 given by (x) = x is continuous
if and only if 1 2 .
Theorem 8.20. Let 1 and 2 be topologies on the same space X .
(i) If 1 2 and 1 is compact, then so is 2 .
(ii) If 1 2 and 2 is Hausdor, then so is 1 .
(iii) If 1 2 , 1 is compact and 2 is Hausdor, then 1 = 2 .
Proof. The routine proof is given on page 64.
The reader may care to recall that Little Bears porridge was neither too
hot nor too cold but just right.
With the hint given by the previous theorem it should be fairly easy to
do do the next exercise.
Exercise 8.21. (i) Give an example of a compact space (X, ) and a topological space (Y, ) together with a continuous bijection f : X Y which is
not a homeomorphism.
(ii) Give an example of a topological space (X, ) and a Hausdor space
(Y, ) together with a continuous bijection f : X Y which is not a homeomorphism.
Solution. See page 64.
We shall give a (not terribly convincing) example of the use of Theorem 8.18 in our proof of Exercise 9.7.
The reader may have gained the impression that compact Hausdor
spaces form an ideal backdrop for continuous functions to the reals. Later
work shows that the impression is absolutely correct but it must be remembered that many important spaces (including the the real line with the usual
topology) are not compact.
23
9
Products of compact spaces
The course contains one further major theorem on compactness.
Theorem 9.1. The product of two compact spaces is compact. (More formally, if (X, ) and (Y, ) are compact topological spaces and is the product
topology, then (X Y, ) is compact.)
Proof. We give a very substantial hint on page 44 and a proof on page 64.
Tychonov showed that the general product of compact spaces is compact (see the note to Exercise 14.10) so Theorem 9.1 is often referred to as
Tychonovs theorem.
The same proof, or the remark that the subspace topology of a product topology is the product topology of the subspace topologies (see Exercise 14.14), gives the closely related result.
Theorem 9.2. Let (X, ) and (Y, ) be topological spaces and let be the
product topology. If K is a compact subset of X and L is a compact subset
of Y , then K L is a compact in .
We know (see Exercise 6.14) that the topology on R2 derived from the
Euclidean metric is the same as the product topology when we give R the
topology derived from the Euclidean metric. Theorem 8.6 thus has the following corollary5 .
Theorem 9.3. Let [a, b] [c, d] be given its usual (Euclidean) topology. Then
the derived topology is compact.
The arguments of the previous section carry over to give results like the
following.
Theorem 9.4. Consider R2 with the standard (Euclidean) topology. A set
E is compact if and only if it is closed and bounded (that is to say, there
exists a M such that x M for all x E ).
5
Stated more poetically by Conway.
If E s closed and bounded, says HeineBorel,
And also Euclidean, then we can tell
That, if it we smother
With a large open cover,
Theres a nite renement as well.
24
Theorem 9.5. Let R2 have the usual metric. If K is a closed and bounded
subset of R2 and f : K R is continuous, then f is bounded and attains its
bounds.
Exercise 9.6. Let R2 have the usual metric. If K is a subset of R2 with
the property that, whenever f : K R is continuous, then f is bounded and
attains its bounds, it follows that K is closed and bounded.
The generalisation to Rn is left to the reader.
The next exercise brings together many of the themes of this course. The
reader should observe that we know what we want the circle to look like.
This exercise checks that dening the circle via quotient maps gives us what
we want.
Exercise 9.7. Consider the complex plane with its usual metric. Let
D = {z C : |z | = 1}
and give D the subspace topology . Give R its usual topology and dene
an equivalence relation by x y if x y Z. We write R/ = T and give
T the quotient topology. The object of this exercise is to show that D and
T are homeomorphic.
(i) Verify that is indeed an equivalence relation.
(ii) Show that, if we dene f : R D by f (x) = exp(2ix), then f (U )
is open whenever U is open.
(iii) If q : R T is the quotient map q (x) = [x] show that q (x) = q (y )
if and only if f (x) = f (y ). Deduce that q f 1 ({exp(2ix)}) = [x] and that
the equation F (exp(2ix)) = [x] gives a well dened bijection F : D T.
(iv) Show that F 1 (V ) = f q 1 (V ) and deduce that F is continuous.
(v) Show that T is Hausdor and explain why D is compact. Deduce
that F is a homeomorphism.
Solution. See page 66.
10
Connectedness
This section deals with a problem which the reader will meet (or has met) in
her rst complex variable course. Here is a similar problem that occurs on the
real line. Suppose that U is an open subset of R (in the usual topology) and
f : U R is a dierentiable function with f (u) = 0 for all u U . We would
like to conclude that f is constant, but the example U = (2, 1) (1, 2),
f (u) = 1 if u > 0, f (u) = 1 if u < 0 shows that the general result is false.
What extra condition should we put on U to make the result true?
25
After some experimentation, mathematicians have come up with the following idea.
Denition 10.1. A topological space (Y, ) is said to be disconnected if we
can nd non-empty open sets U and V such that U V = Y and U V = .
A space which is not disconnected is called connected.
Denition 10.2. If E is a subset of a topological space (X, ) then E is
called connected (respectively disconnected) if the subspace topology on E is
connected (respectively disconnected).
The denition of a subspace topology gives the following alternative characterisation which the reader may prefer.
Lemma 10.3. If E is a subset of a topological space (X, ), then E is disconnected if and only if we can nd open sets U and V such that U V E ,
U V E = , U E = and V E =
Here is another alternative characterisation which shows that we are on
the right track.
Theorem 10.4. If E is a subset of a topological space (X, ), then E is
disconnected if and only if we can nd a non-constant continuous function
f : E R which only takes the value 0 or 1.
Proof. The proof is not hard once you see what is involved. We give the
proof on page 67.
The following deep result is now easy to prove.
Theorem 10.5. If we give R the usual topology, then R and the intervals
[a, b] and (a, b) are connected.
Proof. See page 68.
The reader will nd it instructive to use Theorem 10.4 to prove parts (i)
and (iii) of the next exercise.
Exercise 10.6. (i) If (X, ) and (Y, ) are topological spaces, E is a connected subset of X and g : E Y is continuous then g (E ) is connected.
(More briey the continuous image of a connected set is connected.)
(ii) If (X, ) is a connected topological space and is an equivalence
relation on X , then X/ with the quotient topology is connected.
(iii) If (X, ) and (Y, ) are connected topological spaces, then X Y with
the product topology is connected.
(iv) If (X, ) is a connected topological space and E is a subset of X , it
does not follow that E with the subspace topology is connected.
26
Proof. See page 68.
The proof of the next example is particularly important because it gives
a standard technique for using connectedness in practice.
Example 10.7. Suppose that E is a connected subset of a topological space
(X, ). Suppose that f : E R is locally constant in the sense that, given
any e E , we can nd an open neighbourhood U of e such that f is constant
on U E . Then f is constant.
Proof. There is a hint on page 44 and a proof on page 69.
Exercise 10.8. Suppose that E is subset of a topological space (X, ) such
that any locally constant f : E R is constant. Show that E is connected.
Proof. If you need a hint, look at the proof of Theorem 10.4.
Example 10.7 and Exercise 10.8 together completely settle the question
posed in the rst paragraph of this section.
The following lemma outlines a very natural development.
Lemma 10.9. We work in a topological space (X, ).
(i) Let x0 X . If x0 E and E is connected for all A, then
A E is connected.
(ii) Write x y if there exists a connected set E with x, y E . Then
is an equivalence relation.
(iii) The equivalence classes [x] are connected.
(iv) If F is connected and F [x], then F = [x].
Proof. If you need more details, see page 69.
The sets [x] are known as the connected components of (X, ).
Connectedness is related to another older concept.
Denition 10.10. Let (X, ) be a topological space. We say that x, y X
are path-connected if (when [0, 1] is given its standard Euclidean topology)
there exists a continuous function : [0, 1] X with (0) = x and (1) = y .
Of course, is referred to as a path from x to y .
Lemma 10.11. If (X, ) is a topological space and we write x y if x is
path-connected to y , then is an equivalence relation.
Proof. This just a question of getting the notation under control. We give a
proof on page 70.
27
We say that a topological space is path-connected if every two points in
the space are path-connected.
The following theorem is often useful.
Theorem 10.12. If a topological space is path-connected, then it is connected.
Proof. This is not hard. We give a proof on page 71.
The converse is false (see Example 10.14 below) but there is one very
important case where connectedness implies path-connectedness.
Theorem 10.13. If we give Rn the usual topology, then any open set which
is connected is path-connected.
Proof. We give a hint on page 44 and a proof on page 72.
The following example shows that, even in R2 , we can not remove the
condition open.
Example 10.14. We work in R2 with the usual topology. Let
E1 = {(0, y ) : |y | 1} and E2 = {(x, sin 1/x) : 0 < x 1}
and set E = E1 E2 .
(i) Sketch E .
(ii) Explain why E1 and E2 are path-connected and show that E is closed
and connected.
(iii) Suppose, if possible, that x : [0, 1] E is continuous and x(0) =
(1, 0), x(1) = (0, 0). Explain why we can nd 0 < t1 < t2 < t3 < . . . such
1
that x(tj ) = (j + 2 ) )1 . By considering the behaviour of tj and y (tj ), obtain
a contradiction.
(iv) Deduce that E is not path-connected.
Proof. Parts (ii) to (iv) are done on page 72.
Paths play an important role in complex analysis and algebraic topology.
11
Compactness in metric spaces
When we work in R (or, indeed, in Rn ) with the usual metric, we often use
the theorem of BolzanoWeierstrass that every sequence in a bounded closed
set has a subsequence with a limit in that set. It is also easy to see that
closed bounded sets are the only subsets of Rn which have the property that
every sequence in the set has a subsequence with a limit in that set. This
suggests a series of possible theorems some of which turn out to be false.
28
Example 11.1. Give an example of metric space (X, d) which is bounded
(in the sense that there exists an M with d(x, y ) M for all x, y X ) but
for which there exist sequences with no convergent subsequence.
Solution. We can nd such a space within our standard family of examples.
See page 74.
Fortunately we do have a very neat and useful true theorem.
Denition 11.2. A metric space (X, d) is said to be sequentially compact if
every sequence in X has a convergent subsequence.
Theorem 11.3. A metric space is sequentially compact if and only if it is
compact.
We prove the if and only if parts separately. The proof of the if part is
quite simple when you see how.
Theorem 11.4. If the metric space (X, d) is compact, then it is sequentially
compact.
Proof. There is a hint on page 44 and a proof on page 74
The only if part is more dicult to prove (but also, in my opinion, less
important). We start by proving a result of independent interest.
Lemma 11.5. Suppose that (X, d) is a sequentially compact metric space
and that the collection U with A is an open cover of X . Then there
exists a > 0 such that, given any x X , there exists an (x) A such
that the open ball B (x, ) U(x) .
Proof. There is a hint on page 45 and a proof on page 74.
We now prove the required result.
Theorem 11.6. If the metric space (X, d) is sequentially compact, it is compact.
Proof. There is a hint on page 45 and a proof on page 75.
This gives an alternative but less instructive proof of the theorem of
HeineBorel.
Alternative proof of Theorem 8.5. By the BolzanoWeierstrass theorem, [a, b]
is sequentially compact. Since we are in a metric space, it follows that [a, b]
is compact.
29
If you prove a theorem on metric spaces using sequential compactness it is
good practice to try and prove it directly by compactness. (See, for example,
Exercise 14.19.)
The reader will hardly need to be warned that this chapter dealt only
with metric spaces. Naive generalisations to general topological spaces are
likely to be meaningless or false.
12
The language of neighbourhoods
One of the lines of thought involved in the birth of analytic topology was
initiated by Riemann. We know that many complicated mathematical structures can be considered as a space which locally looks like a simpler space.
Thus the surface of the globe we live on is sphere but we consider it locally as
a plane (ie like R2 ). The space we live in looks locally like R3 but its global
structure could be very dierent. For example, Riemann says Space would
necessarily be nite if . . . [we] ascribed to it a constant curvature, as long as
that curvature had a positive value, however small. [Riemanns discussion
On the Hypotheses which lie at the Foundations of Geometry is translated
and discussed in the second volume of Spivaks Dierential Geometry.]
Unfortunately the mathematical language of his time was not broad
enough to allow the expression of Riemanns insights. If we are given a
particular surface such as sphere, it is easy, starting with the complete structure, to see what locally and resembles might mean, but, in general, we
seem to be stuck in a vicious circle with locally only meaningful when the
global structure is known and the global structure only knowable when the
meaning of locally is known.
The key to the problem was found by Hilbert who, in the course of his
investigations into the axiomatic foundations of geometry produced an axiomatisation of the notion of neighbourhood in the Euclidean plane R2 . By
developing Hilberts ideas, Weyl obtained what is essentially the modern
denition of a Riemann surface (this object, which looks locally like C, was
another brilliant creation of Riemann).
However, although the notion of an abstract space with an abstract notion
of closeness was very much in the air, there were a large number of possible
candidates for such an abstraction. It was the achievement of Hausdor to
see in Hilberts work the general notion of a neighbourhood.
Although Hausdor dened topologies in terms of neighbourhoods, it
appears to be technically easier to dene topologies in terms of open sets
as we have done in this course. However, topologists still use the notion of
neighbourhoods.
30
We have already dened an open neighbourhood of x to be an open set
containing x. We now give the more general denition.
Denition 12.1. Let (X, ) be a topological space. If x X we say that N
is a neighbourhood of x if we can nd U with x U N .
The reader may check her understanding by proving the following easy
lemmas.
Lemma 12.2. Let (X, ) be a topological space. Then U if and only if,
given x U , we can nd a neighbourhood N of x with N U .
Proof. The easy proof is given on page 75.
Lemma 12.3. Let (X, ) and (Y, ) be topological spaces. Then f : X Y
is continuous if and only if, given x X and M a neighbourhood of f (x) in
Y , we can nd a neighbourhood N of x with f (N ) M .
Proof. The easy proof is given on page 76.
Exercise 12.4. (i) If (X, d) is a metric space, show that N is a neighbourhood of x if and only we can nd an > 0 such that the open ball B (x, ) N .
(ii) Consider R with the usual topology. Give an example of a neighbourhood which is not an open neighbourhood. Give an example of an unbounded
neighbourhood. Give an example of a neighbourhood which is not connected.
We end the course with a warning. Just as it is possible to dene continuous functions in terms of neighbourhoods so it is possible to dene convergence in terms of neighbourhoods. This works well in metric spaces.
Lemma 12.5. If (X, d) is a metric space, then xn x, if and only if given N
a neighbourhood of x, we can nd an n0 (depending on N ) such that xn N
for all n n0 .
However, things are not as simple in general topological spaces.
Denition 12.6. [WARNING. Do not use this denition without
reading the commentary that follows.] Let (X, ) be a topological space.
If xn X and x X then we say xn x if, given N a neighbourhood of x
we can nd n0 (depending on N ) such that xn N for all n n0 .
Any hopes that limits of sequences will behave as well in general topological spaces are dashed by the following example.
Example 12.7. Let X = {a, b} with a = b. If we give X the indiscrete
topology, then, if we set xn = a for all n, we have xn a and xn b.
31
Thus limits need not be unique.
Of course, it is possible to persist in spite of this initial shock, but the
reader will nd that she can not prove the links between limits of sequences
and topology that we would wish to be true. This failure is not the readers
fault. Deeper investigations into set theory reveal that sequences are inadequate tools for the study of topologies which have neighbourhood systems
which are large in the set theoretic sense. (Exercise 14.25 represents an
attempt to show what this means.) It turns out that the deeper study of set
theory reveals not only the true nature of the problem but the solution via
nets (a kind of generalised sequence) or lters (preferred by the majority of
mathematicians).
13
Books
If the reader looks at a very old book on general (or analytic ) topology, she
may nd both the language and the contents rather dierent from what she is
used to. In 1955 Kelley wrote a book General Topology [1] which stabilised
the content and notation which might be expected in advanced course on
the subject. Texts like [3] (now in a very cheap Dover reprint6 ) and [2]
which extracted a natural elementary course quickly appeared and later texts
followed the established pattern. Both [3] and [2] are short and sweet. With
luck, they should be in your college library. The book of Sutherland [4] has
the possible advantage of being written for a British audience and the certain
advantage of being in print7 .
Many books on Functional Analysis, Advanced Analysis, Algebraic Topology and Dierential Geometry cover the material in this course and then go
on to develop it in the directions demanded by their particular subject.
References
[1] Kelley, J. L, General Topology, Princeton N. J,. Van Nostrand, 1955.
[2] Manseld, M. J., Introduction to Topology, Princeton N. J., Van Nostrand, 1963.
[3] Mendeleson, B., Introduction to Topology, Boston Mass., Allyn and Bacon, 1962. [Now available in a Dover reprint, New York, Dover, 1990]
6
7
October, 2004.
October, 2004.
32
[4] Sutherland W. A., Introduction to Metric and Topological Spaces, Oxford,
OUP, 1975.
14
Exercises
Exercise 14.1. Let X be a set and d : X 2 R a function with the following
properties.
(i) d(x, x) = 0 for all x X .
(ii) d(x, y ) = 0 implies x = y .
(iv) d(y, x) + d(y, z ) d(x, z ) for all x, y, z X .
Show that d is a metric on X .
Exercise 14.2. (i) If V is an inner product space and
is the standard
norm derived from the inner product, prove the parallelogram law
a+b
2
+ ab
2
= 2( a
2
+ b 2 ).
(ii) Give an example of a normed vector space where their norm can not
be derived from an inner product in a standard way.
Exercise 14.3. Let RN have its usual (Euclidean) metric.
(i) Suppose that fj : Rnj Rmj is continuous for 1 j k . Show that
the map f : Rn1 +n2 ++nk Rm1 +m2 ++mk given by
f (x1 , x2 , . . . , xk ) = (f1 (x1 ), f2 (x2 ), . . . , fk (xk ))
is continuous.
(ii) Show that the map U : Rn Rkn given by
U (x) = (x, x, . . . , x)
is continuous.
(iii) Suppose that gj : Rn Rmj is continuous for 1 j k . Use the
composition law to show that the map g : Rn Rm1 +m2 ++mk given by
g (x) = (g1 (x), g2 (x), . . . , gk (x))
is continuous.
(iv) Show that the map A : R2 R given by A(x, y ) = x + y is continuous.
(v) Use the composition law repeatedly to show that the map f : R2 R
given by
xy
f (x, y ) = sin
2 + y2 + 1
x
is continuous.
[If you have diculty with (v) try smaller subproblems. For example, can
you show that (x, y ) x2 + y 2 is continuous?]
33
Exercise 14.4. Consider R with the ordinary Euclidean metric.
(i) We know that sin : R R is continuous. Show that if U = R, then
U is open but sin U is not.
(ii) We dene a function f : R R as follows. If x R set x = x [x]
and write
x = .x1 x2 x3 . . .
as a decimal, choosing the terminating form in case of ambiguity. If x2n+1 =
0 for all suciently large n, let N be the least integer such that x2n+1 = 0
for all n N , and set
N
x2N +2j 10N j .
f (x) = (1)
j =0
We set f (x) = 0 otherwise.
Show that if U is a non-empty open set, f (U ) = R and so f (U ) is open
but that f is not continuous.
Exercise 14.5. Let (X, d) be a metric space and let r > 0. Show that
B (x, r) = {y : d(x, y ) r}
is a closed set:(a) By using the denition of a closed set in terms of limits.
(b) By showing that the complement of B (x, r) is open.
We call B (x, r) the closed ball centre x and radius r.
Exercise 14.6. Prove Theorems 4.6 and 4.7 directly from the denition of
a closed set in terms of limits without using open sets.
Exercise 14.7. (i) Let (X, d) be a metric space. Show that
d(x, y )
1 + d(x, y )
(x, y ) =
denes a new metric on X .
(ii) Show that, in (i), d and have the the same open sets.
(iii) Suppose that d1 , d2 , . . . are metrics on X . Show that
(x, y ) =
n=1
2n dn (x, y )
)
1 + dn (x, y )
denes a metric on X .
34
Exercise 14.8. (This is just intended to remind of you of some elementary
results on maps.) Let X and Y be non-empty sets and f : X Y be a
function. Suppose that A, A X , B, B Y , A X , B Y for all
. Which of the following statements are always true and which may be
false? Give a counter example or a brief explanation in each case.
(i) f ( A ) = f (A ).
(ii) f ( A ) = f (A ).
(iii) f (A \ A ) = f (A) \ f (A ).
(iv) f 1 ( B ) = f 1 (B ).
(v) f 1 ( B ) = f 1 (B ).
(vi) f 1 (B \ B ) = f 1 (B ) \ f 1 (B ).
How would your answers change if f was bijective?
Exercise 14.9. Let (X, ) be a topological space.
(i) Show that, if E is subset of X , there exists a unique open set V such
that
(a) V E ,
(b) if U is a open set with U E , then U V .
We call V the interior of E and write Int E = V .
(ii) Show by means of an example that the following statement may be
false.
If E is subset of X , there exists a unique open set V such that
(a) V E ,
(b) if U is a open set with U E , then U V .
(iii) Show that, if E is subset of X , there exists a unique closed set F
such that
(a) F E ,
(b) if G is a closed set with G E , then G F .
We call F the closure of E and write Cl E = F .
(iv) Show that X \ Int E = Cl(X \ E ).
(v) If is derived from a metric d, show that the closure of E consists
precisely of those points x X such that there exists a sequence of points en
in E with d(en , x) 0 as n .
Exercise 14.10. (i) Suppose that A is non-empty and that (X , ) is a
topological space. Explain what is meant by saying that is the smallest
topology on A X for which each of the projection maps : A X
X is continuous and explain why we know that it exists. We call the
product topology.
35
(ii) Show that U if and only if, given x U , we can nd U
[ A] such that
x
U
A
and U = X for all but nitely many of the .
(iii) By considering A = [0, 1] and taking each (X , ) to be a copy of R
show that the following condition denes a topology on the space R[0,1] of
functions f : [0, 1] R. A set U if and only if, given any f0 U , there
exists an > 0 and x1 , x2 , . . . , xn [0, 1] such that
{f R[0,1] : |f (xj ) f0 (xj )| < for all 1 j n} U.
[The reader who can not see the point of this topology is in good, but mistaken,
company. The great topologist Alexandrov recalled that when Tychonov (then
aged only 20) produced this denition His chosen . . . denition seemed not
only unexpected but perfectly paradoxical. [I remember] with what mistrust
[I] met Tychonovs proposed denition. How was it possible that a topology
induced by means of such enormous neighbourhoods, which are only distinguished from the whole space by a nite number of the coordinates, could
catch any of the essential characteristics of a topological product? However,
Tychonovs choice was justied by its consequences, in particular, the generalisation (by Tychonov) of Theorem 9.1 to show that the (Tychonov) product
of compact spaces is compact. This theorem called Tychonovs theorem is
one of the most important in modern analysis.
In common with many of the most brilliant members of the Soviet school,
Tychonov went on to work in a large number of branches of pure and applied
mathematics. His best known work includes a remarkable paper on solutions
of the heat equation8 .]
Exercise 14.11. (i) Let X = {a, b} with a = b. Show that there does not
exist a largest topology contained in = {, {a}, {b}, X }. (More formally,
show that there does not exist a topology on X such that and such
that, if is any topology with , then .) Compare and contrast
Lemma 6.5.
(ii) (with Show the notation of Lemma 6.15) that the quotient topology
on X/ is the largest topology (in the sense of (i)) such that q : X X/
is continuous.
Exercise 14.12. Consider R with the usual (Euclidean) topology. Let x y
if and only if x y Q. Show that is an equivalence relation. Show that
8
A substantial part of Volume 22, Number 2 of Russian Mathematical Surveys 1967 is
devoted to Tychonov and his work. The quotation from Alexandrov is taken from there.
36
R/ is uncountable but that the quotient topology on R/ is the indiscrete
topology.
Exercise 14.13. (i) If (X, ) is a topology derived from a metric show that,
given x X , we can nd open sets Uj [1 j ] such that {x} = Uj .
j =1
(ii) Show, by verifying the conditions for a topological space directly (so
you may not quote Exercise 14.10, that the following condition denes a
topology on the space R[0,1] of functions f : [0, 1] R. A set U if and
only if, given any f0 U , there exists an > 0 and x1 , x2 , . . . , xn [0, 1]
such that
{f R[0,1] : |f (xj ) f0 (xj )| < for 1 j n} U.
(iii) Show that the topology is Hausdor but can not be derived from a
metric.
Exercise 14.14. Let (X, ) and (Y, ) be topological spaces with subsets E
and F . Let the subspace topology on E be E and the subspace topology on F
be F . Let the product topology on X Y derived from and be and let
the product topology on E F derived from E and F be . Show that is
the subspace topology on E F derived from .
Exercise 14.15. (i) Let Hi be a collection of subsets of Xi and let i be the
smallest topology on Xi containing Hi [i = 1, 2]. If f : X1 X2 has the
property that f 1 (H ) H1 whenever H H2 , show that f is continuous
(with respect to the topologies 1 and 2 ).
(ii) Suppose that (X, ) and (Y, ) are topological space and we give X Y
the product topology. If (Z, ) is a topological space, show that f : Z X Y
is continuous if and only if X f : Z X and Y f : Z Y are
continuous.
(iii) Let R have the usual topology (induced by the Euclidean metric) and
let R2 have the product topology (which we know to be the usual topology
induced by the Euclidean metric). Dene
f (x, y ) =
xy
x2 +y 2
if (x, y ) = (0, 0),
if (x, y ) = (0, 0).
0
Show that, if we dene hx (y ) = gy (x) = f (x, y ) for all (x, y ) R2 , then
the function hx : R R is continuous for each x R and the function
gy : R R is continuous for each y R. Show, however, that f is not
continuous.
37
Exercise 14.16. In complex variable theory we encounter uniform convergence on compacta. This question illustrates the basic idea basic in the case
of C () the space of continuous functions f : C where
= {z C : |z | < 1}.
(i) Show, by means of an example, that an f C () need not be bounded
on C ().
(ii) Explain why
dn (f, g ) =
sup
|f (z ) g (z )|
|z |11/n
exists and is nite for each n 1 and all f, g C (). Show that dn satises
the triangle law and symmetry but give an example of a pair of functions
f, g C () with f = g yet dn (f.g ) = 0.
(iii) Show that
2n dn (f, g )
d(f, g ) =
1 + dn (f, g )
n=1
exists and is nite for all f, g C ().
(iv) Show that d is a metric on C ().
[If you require a hint, do Exercise 14.7 (i).]
Exercise 14.17. [This requires Exercise 14.9.]
(i) Show that the closure of a connected set is connected.
(ii) Deduce that connected components are closed.
(iii) Show that if there are only nitely many components they must all
be open.
(iv) Find the connected components of
{0}
{1/n : n 1, n Z}
with the usual metric.
Which are open and which are not? Give reasons.
Exercise 14.18. (i) If (X, ) and (Y, ) are topological spaces, E is a pathconnected subset of X and g : E Y is continuous, show that g (E ) is
path-connected. (More briey the continuous image of a path-connected set
is path-connected.)
(ii) If (X, ) is a path-connected topological space and is an equivalence
relation on X , show that X/ with the quotient topology is path-connected.
(iii) If (X, ) and (Y, ) are path-connected topological spaces, show that
X Y with the product topology is path-connected.
38
(iv) If (X, ) is a path-connected topological space and E is a subset of
X , show that it does not follow that E with the subspace topology is pathconnected.
Exercise 14.19. Suppose that (X, d) is a compact metric space, (Y, ) is a
metric space and f : X Y is continuous. Explain why, given > 0, we
can nd, for each x X , a x > 0 such that, if d(x, y ) < 2x , it follows
that (f (x), f (y )) < /2. By considering the open cover B (x, x ) and using
compactness, show that there exists a > 0 such that d(x, y ) < implies
(f (x), f (y )) < . (In other words, a continuous function from a compact
metric space to a metric space is uniformly continuous.)
Exercise 14.20. Which of the following spaces are homeomorphic and which
are not? Give reasons.
(i) R with the usual topology.
(ii) R with the discrete topology.
(iii) Z with the discrete topology.
(iv) [0, 1] with the usual topology.
(v) (0, 1) with the usual topology.
[This is rather feeble question but in this short course we have not found
enough topological properties to distinguish between some clearly distinguishable topological spaces. We return to this matter in Exercise 14.21.]
Exercise 14.21. Suppose that f : [0, 1] R and g : [0, 1] R are continuous maps with f (0) = 1, f (1) = 2, g (0) = 0 and g (1) = 1. Show
that
f ([0, 1]) g ([0, 1]) =
(In other words, the two paths must cross.)
Show that R and R2 with the usual topologies are not homeomorphic. Are
[0, 1] and the circle
{z C : |z | = 1}
homeomorphic and why?
[But are R2 and R3 homeomorphic? Questions like this form the beginning
of modern algebraic topology.]
Exercise 14.22. Which of the following statements are true and which false.
Give a proof or counter-example.
(i) If a topological space (X, ) is connected then the only sets which are
both open and closed are X and .
(ii) If every set in a topological space (X, ) is open or closed (or both)
then is the discrete topology.
39
(iii) Every open cover of R with the usual topology has a countable subcover.
(iv) Suppose that and are topologies on a space X with . Then,
if (X, tau) is connected, so is (X, ).
(v) Suppose that and are topologies on a space X with . Then,
if (X, ) is connected, so is (X, ).
Exercise 14.23. [Bases of neighbourhoods.](i) Let (X, ) be a topological
space. Write Nx for the set of neighbourhoods of x X . Prove the following
results.
(1) Nx = .
(2) If N Nx , then x N .
(3) If N, M Nx , then N M Nx .
(4) If N Nx and M N then M Nx .
(5) If N Nx then there exists an U Nx such that U N and
U Ny for all y U .
(ii) Suppose that X is a set such that each x X is associated with a
collection Nx of subsets of X . If conditions (1) to (4) of part (ii) hold, show
that the family of sets U such that, if x U , then we can nd an N Nx
with N U is a topology on X . If, in addition, condition (5) holds show
that Nx is the collection of -neighbourhoods of x for each x X .
Exercise 14.24. Consider R2 with the usual Euclidean topology. Let
E = {(x, 1) : x R} {(x, 1) : x R}
and give E the subspace topology.
Dene a relation on E by taking
(x, y ) (x, y )
(x, y ) (x, y )
for all (x, y ) E
for all (x, y ) E with x = 0.
Show that that is an equivalence relation on E .
Now give E/ the equivalence relation. Show that if [(x, y )] E/ we
can nd an open neighbourhood U of [(x, y )] which is homeomorphic to R.
Show, however, that E/ is not Hausdor.
[This nasty example shows that looks nice locally is not sucient to give
looks nice globally. It is good start to a course in dierential geometry to
ask what extra conditions are required to make sure that a space that looks
locally like a line looks globally like a line or a circle.]
Exercise 14.25. Consider the collection X of all functions f : [0, 1] R
with f (x) > 0 for x > 0, f (0) = 0 and f (x) 0 as x 0. We take
40
X = X {f0 } where f0 is the zero function dened by f0 (x) = 0 for all
x [0, 1]. If g X , write
Ug = {f X : f (x)/g (x) 0 as x 0}.
Show that, given g1 , g2 X we can nd a g3 X such that
Ug3 Ug1 Ug2 .
Conclude that, if consists of together with all those sets V such that
V Ug for some g X , then is a topology on X . Show that
Ug = {f0 }
g X
Now suppose gj X . If we set g (0) = 0 and
g (t) = n1 min gj (t) for t (n + 1)1 , n1 ,
1j n
show that g X and gj Ug . Conclude that, although every open neigh/
bourhood of f0 contains innitely many points and the intersection of the
open neighbourhoods of f0 is the one point set {f0 }, there is no sequence gj
with gj = f0 such that gj f0 .
[If you just accept this result without thought it is not worth doing the
question. You should compare and contrast the metric case. I would say that
f0 is surrounded by too many neighbourhood-shells to be approached by a
sequence but the language of the course is inadequate to make this thought
precise.
I am told that the ancient Greek geometers used a similar counter example
for a related purpose.]
Exercise 14.26. The object of this exercise is to produce an interesting metric (the Poincar metric) associated with complex analysis.
e
(i) Let D = {z C : |z | < 1} Show that, if |a| < 1 the Mbius map Ta
o
given by
az
Sa (z ) =
1 a z
maps D to D and interchanges the points 0 and 1.
(ii) Show that the only Mbius maps which take D to D and x 0 are
o
rotations.
(iii) Show that the collection G of Mbius maps which take D to D is a
o
subgroup of M.
41
(iv) Show that the elements of G are compositions of rotations and maps
of the form Sa .
(iv) Suppose that we seek a metric d on D which is invariant under Mbius
o
maps S which take D to D (i.e. such that d(Sz1 , Sz2 ) = d(z1 , z2 ) for all
z1 , z2 C and S G ). Show that, if we write f (x) = d(x, 0) for x real and
0 x < 1, then
z1 z
d(z1 , z2 ) = f
1 a z
(v) Let us guess that d(0, x) + d(x, y ) = d(0, y ) whenever x and y are
real and 0 x y and that f : [0, 1) R is everywhere dierentiable (in
particular, right dierentiable at 0). Show that
f (x) =
Deduce that
d(z1 , z2 ) = A log
f (0)
.
1 x2
|1 z1 z2 | + |z1 z2 |
|1 z1 z2 | |z1 z2 |
for all z1 , z2 D and some real constant A > 0.
(vi) We still have to show that
(z1 , z2 ) = log
|1 z1 z2 | + |z1 z2 |
|1 z1 z2 | |z1 z2 |
does, in fact, give a well dened metric on D which is invariant under Mbius
o
maps which take D to D. You should nd is straightforward to prove all of
these statements with the possible exception of the triangle inequality.
(vii) Here is one way to prove the triangle inequality. (I make no claim
that it is the best.) First show, by taking a large sheet of paper and doing the
algebra, that, if R, r and are real and 1 > R, r 0, then
(R + r)2 |1 Rrei |2 (1 + Rr)2 |R rei |2 0.
Deduce that, if z1 , z2 D, then
z1 z2
|z1 | + |z2 |
.
1 + |z1 ||z2 |
1 z1 z2
Hence show that
log
1 + |z1 |
1 |z1 |
+ log
1 + |z2 |
1 |z2 |
42
log
1+
z1 z2
1z1 z2
1
z1 z2
1z1 z2
and deduce the triangle inequality.
[This proof is not very illuminating and the usual approach to the Poincar
e
metric is through geodesics.]
(viii) Show that given a D and r > 0 we can nd b D and R > 0
such that
{z : (z, a) = r} = {z : |z b| = R}.
Sketch {z : (z, a) = r} for a sequence of r .
(xi) Let H = {z C : z > 0. Show that writing
d(z1 , z2 ) = log
|z1 z2 | + |z1 z2 |
|z1 z2 | |z1 z2 |
gives a well dened metric d on H which is invariant under Mbius maps
o
which take H to H .
15
Some hints
Theorem 8.5. [The HeineBorel Theorem.] Let R be given its usual
(Euclidean) topology. Then the closed bounded interval [a, b] is compact.
Hint. Suppose that C is an open cover of [a, b]. If C1 is a nite subcover of
[a, c] and C2 is a nite subcover of [c, a], then C1 C2 is a nite cover of [a, b].
We can use this as a basis for a lion hunting (bisection) argument.
[Return to page 20 or go to a full proof on 60.]
Theorem 8.8. If (X, ) is Hausdor, then every compact set is closed.
Hint. Let K be a compact set. If x K , then, given any k K , we know
/
that k = x and so, since X is Hausdor, we can nd open sets Uk and Vk
such that
k Vk , x Uk and Vk Uk = .
Now use compactness.
[Return to page 21 or go to a full proof on 61.]
Theorem 8.18. Let (X, ) be a compact and (Y, ) a Hausdor topological
space. If f : X Y is a continuous bijection, then it is a homeomorphism.
Hint. Observe that we need only show that f (K ) is closed whenever K is
closed.
[Return to page 23 or go to a full proof on 63.]
43
Theorem 9.1. The product of two compact spaces is compact. (More formally, if (X, ) and (Y, ) are compact topological spaces and is the product
topology, then (X Y, ) is compact.)
Hint. Let {O }A be an open cover for X Y . Then given (x, y ) X Y
we can nd Ux,y , Vx,y and (x, y ) A such that
(x, y ) Ux,y Vx,y O(x,y) .
Now show that, for each x X , we can nd a positive integer n(x) and
y (x, j ) Y [1 j n(x)] such that
n(x)
Vx,y(x,j ) = Y.
j =1
(x
Now consider the Ux = n=1) Ux,y(x,j ) .
j
[Return to page 24 or go to a full proof on 64.]
Example 10.7. Suppose that E is a connected subset of a topological space
(X, ). Suppose that f : E R is locally constant in the sense that, given
any e E , we can nd an open neighbourhood U of e such that f is constant
on U E . Then f is constant.
Hint. There is no loss in generality in taking E = X and X = . Choose an
x0 X and set c = f (x0 ). Now consider the sets
U = {x X : f (x) = c} and V = {x X : f (x) = c}.
[Return to page 27 or go to a full proof on 69.]
Theorem 10.13. If we give Rn the usual topology, then any open set which
is connected is path-connected.
Hint. Pick x and let U be the set of all points in which are pathconnected to x and let V be the set of all points in which are not. We
need to prove that U and V are open and to do this we make use of the fact
that any point in an open ball is path-connected to the centre of the ball.
[Return to page 28 or go to a full proof on page 72.]
Theorem 11.4. If the metric space (X, d) is compact it is sequentially compact.
44
Hint. Let xn be a sequence in X . If it has no convergent subsequence then,
for each x X , we can nd a (x) > 0 and an N (x) such that xn B (x, (x))
/
for all n N (x).
[Return to page 29 or go to a full proof on page 74.]
Lemma 11.5. Suppose that (X, d) is a sequentially compact metric space
and that the collection U with A is an open cover of X . Then there
exists a > 0 such that, given any x X , there exists an (x) A such
that the open ball B (x, ) U(x) .
Hint. Suppose the rst sentence is true and the second sentence false. Then,
for each n 1, we can nd an xn such that B (xn , 1/n) U for all A.
[Return to page 29 or go to a full proof on page 74.]
Theorem 11.6. If the metric space (X, d) is sequentially compact, it is compact.
Hint. Let (U )A be an open cover and let be dened as in Lemma 11.5.
The B (x, ) form a cover of X . If they have no nite subcover then, given
x1 , x2 , . . . xn , we can nd an xn+1 n=1 B (xj , ).
/j
[Return to page 29 or go to a full proof on page 75.]
16
Some proofs
Exercise 2.2. If d : X 2 R is a function with the following properties:(ii) d(x, y ) = 0 if and only if x = y ,
(iii) d(x, y ) = d(y, x) for all x, y X ,
(iv) d(x, y ) + d(y, z ) d(x, z ) for all x, y, z X ,
show that d is a metric on X .
Proof. Setting z = x in condition (iv) and using (iii) and (ii), we have
2d(x, y ) = d(x, y ) + d(y, x) d(x, x) = 0
so d(x, y ) 0.
[Return to page 3.]
Exercise 2.4. Let X = {a, b, c} with a, b and c distinct. Write down functions dj : X 2 R satisfying condition (i) of Denition 2.1 such that
(1) d1 satises conditions (ii) and (iii) but not (iv).
(2) d2 satises conditions (iii) and (iv) but it is not true that x = y
implies d(x, y ) = 0.
45
(3) d3 satises conditions (iii) and (iv) and x = y implies d3 (x, y ) = 0.
but it is not true that d3 (x, y ) = 0 implies x = y .
(4) d4 satises conditions (ii) and (iv) but not (iii).
You should verify your statements.
Solution. Here are some possible choices.
(1) Take d1 (x, x) = 0 for all x X , d1 (a, b) = d1 (b, a) = d1 (a, c) =
d1 (c, a) = 1 and d1 (b, c) = d1 (c, b) = 3. Conditions (ii) and (iii) hold by
inspection but
d1 (b, a) + d1 (a, c) = 2 < 3 = d1 (b, c).
(2) Take d2 (x, x) = 1 and d2 (x, y ) = 2 if x = y . Condition (ii) fails and
condition (iii) holds by inspection. We observe that
d2 (x, y ) + d2 (y, z ) 1 + 1 = 2 d2 (x, z )
so the triangle law holds.
(3) Take d2 (x, y ) = 0 for all x, y X .
(4) Take d4 (x, x) = 0 for all x X , d4 (a, b) = d4 (b, a) = d4 (a, c) =
d4 (c, a) = 1 and d1 (b, c) = d1 (c, b) = 5 . Conditions (ii) holds, and condi4
tion (iii) fails by inspection and
d(x, y ) + d(y, z ) = d(x, y ) = d(x, z ) d(x, z )
d(x, y ) + d(y, z ) = d(y, z ) = d(x, z ) d(x, z )
5
d(x, y ) + d(y, z ) 1 + 1 = 2 4 d(x, z )
if y = z,
if x = y,
otherwise,
so the triangle law holds.
[Return to page 4.]
Exercise 2.8. If (V,
) is a normed vector space, then the condition
d(u, v) = u v
denes a metric d on V .
Proof. We observe that
d(u, v) = u v 0
and
d(u, u) = 0 = 00 = |0| 0 = 0 0 = 0.
Further, if d(u, v) = 0, then u v = 0 so u v = 0 and u = v. We also
observe that
d(u, v) = u v = (1)(v u) = | 1| v u = d(v, u)
46
and
d(u, v) + d(v, w) = u v + v w
(u v) + (v w)
= u w = d(u, w).
[Return to page 5.]
Lemma 2.10. The the discrete metric on X is indeed a metric.
Proof. The only non-evident condition is the triangle law. But
d(x, y ) + d(y, z ) = d(x, y ) = d(x, z ) d(x, z )
d(x, y ) + d(y, z ) = d(y, z ) = d(x, z ) d(x, z )
d(x, y ) + d(y, z ) 1 + 1 = 2 1 d(x, z )
if y = z,
if x = y,
otherwise,
so we are done.
[Return to page 5.]
Lemma 2.16. Show that the British Rail express metric and the British Rail
stopping metric are indeed metrics.
Solution. We show that the British Rail stopping metric is indeed a metric.
The case of the British Rail express metric is left to the reader.
Let d be the British rail stopping metric on R2 . It is easy to see that
d(u, v) 0 and that d(u, v) = d(v, u). Since u and u are linearly dependent,
d(u, u) = u u
2
=0
2
= 0.
If d(u, v) = 0, then we know that at least one of the following statements
is true
(1) u v 2 = 0 and so u v = 0,
(2) u 2 + v 2 = 0 and so u 2 = v 2 = 0 whence u v = 0.
In either case u = v as required.
It only remains to prove the triangle inequality. Observe that, if v and
w are not linearly dependent,
d(u, v) + d(v, w) u v
2
+v
2
+w
2
u
By similar reasoning
d(u, v) + d(v, w) d(u, w)
47
2
+w
2
d(u, w).
if u and v are not linearly dependent. Finally, if u and v are linearly dependent and v and w are linearly dependent, then u and w are linearly
dependent so
d(u, v) + d(v, w) = u v
2
+ vw
2
uw
2
= d(u, w).
Thus the triangle law holds.
[Return to page 7.]
Lemma 3.3. [The composition law.] If (X, d) and (Y, ) and (Z, ) are
metric spaces and g : X Y , f : Y Z are continuous, then so is the
composition f g .
Proof. Let > 0 be given and let x X . Since f is continuous, we can nd
a 1 > 0 (depending on and f g (x) = f (g (x)) with
(f (g (x)), f (y )) < whenever (g (x), y ) < 1 .
Since g is continuous, we can nd a 2 > 0 such that
(g (x), g (t)) < 1 whenever d(x, t) < 2 .
We now have
(f (g (x)), f (g (t))) < whenever d(x, t) < 2
as required.
[Return to page 7.]
Exercise 3.4. Let R and R2 have their usual (Euclidean) metric.
(i) Suppose that f : R R and g : R R are continuous. Show that the
map (f, g ) : R2 R2 is continuous.
(ii) Show that the map M : R2 R given by M (x, y ) = xy is continuous.
(iii) Use the composition law to show that the map m : R2 R given by
m(x, y ) = f (x)g (y ) is continuous.
Solution. (i) Let (x, y ) R2 . Given > 0, we can nd 1 > 0 such that
|f (x) f (s)| < /2 whenever |x s| < 1
and 2 > 0 such that
|g (y ) g (t)| < /2 whenever |y t| < 2 .
48
If we set = min(1 , 2 ), then (x, y ) (s, t)
2
< implies
|x s| < 1 and |y t| < 2
so that
|f (x) f (s)| < /2 and |g (y ) g (t)| < /2
whence
(f (x), g (y )) (f (s), g (t))
2
(f (x), 0) (f (s), 0) 2 + (0, g (y )) (0, g (t))
= |f (x) f (s)| + |g (y ) g (t)| <
as required.
[Return to page 7.]
Example 3.6. (i) Let (X, d) be a metric space. If r > 0, then
B (x, r) = {y : d(x, y ) < r}
is open.
(ii) If we work in Rn with the Euclidean metric, then the one point set
{x} is not open.
(iii) If (X, d) is a discrete metric space, then
{x} = B (x, 1/2)
and all subsets of X are open.
Proof. (i) If y B (x, r), then = r d(x, y ) > 0 and, whenever d(z, y ) < ,
the triangle inequality gives us
d(x, z ) d(x, y ) + d(y, z ) < r
so z B (x, r). Thus B (x, r) is open.
(ii) Choose e Rn with e 2 = 1. (We could take e = (1, 0, 0, . . . , 0).)
If > 0, then, setting y = x + (/2)e, we have x y 2 < , yet y {x}.
/
Thus {x} is not closed.
(iii) Observe that d(x, x) = 0 < 1/2 and d(x, y ) = 1 > 1/2 for x = y . If
x E then d(x, y ) < 1/2 implies y = x E so E is open.
[Return to page 9.]
Theorem 3.7. If (X, d) is a metric space, then the following statements are
true.
(i) The empty set and the space X are open.
(ii) If U is open for all A, then A U is open. (In other words,
the union of open sets is open.)
(iii) If Uj is open for all 1 j n, then n=1 Uj is open.
j
49
2
Proof. (i) Since there are no points e in , the statement
x whenever d(x, e) < 1
holds for all e . Since every point x belongs to X , the statement
x X whenever d(x, e) < 1
holds for all e X .
(ii) If e A U , then we can nd a particular 1 A with e U1 .
Since U1 is open, we can nd a > 0 such that
x U1 whenever d(x, e) < .
Since U1
A
U ,
x
U whenever d(x, e) < .
A
Thus A U is open.
(iii) If e n=1 Uj then e Uj for each 1 j n. Since Uj is open, we
j
can nd a j > 0 such that
x Uj whenever d(x, e) < j .
Setting = min1j n j , we have > 0 and
x Uj whenever d(x, e) <
for all 1 j n. Thus
n
x
Uj whenever d(x, e) <
j =1
and we have shown that
[Return to page 9.]
n
j =1
Uj is open.
Theorem 3.9. Let (X, d) and (Y, ) be metric spaces. A function f : X Y
is continuous if and only if f 1 (U ) is open in X whenever U is open in Y .
Proof. Suppose rst that f is continuous and that U is open in Y . If x
f 1 (U ), then we can nd a y U with f (x) = y . Since U is open in Y , we
can nd an > 0 such that
z U whenever (y, z ) < .
50
Since f is continuous, we can nd a > 0 such that
(y, f (w)) = (f (x), f (w)) < whenever d(x, w) < .
Thus
f (w) U whenever d(x, w) < .
In other words,
w f 1 (U ) whenever d(x, w) < .
We have shown that f 1 (U ) is open.
We now seek the converse result. Suppose that f 1 (U ) is open in X
whenever U is open in Y . Suppose x X and > 0. We know that the open
ball
B (f (x), ) = {y Y : (f (x), y ) < }
is open. Thus x f 1 B f (x), and f 1 B f (x), is open. It follows that
there is a > 0 such that
w f 1 B f (x), whenever d(x, w) < ,
so, in other words,
(f (x), f (w)) < whenever d(x, w) < .
Thus f is continuous.
[Return to page 9.]
Example 3.10. Let X = R and d be the discrete metric. Let Y = R and
be the usual (Euclidean) metric.
(i) If we dene f : X Y by f (x) = x, then f is continuous but there
exist open sets U in X such that f (U ) is not open.
(ii) If we dene g : Y X by g (y ) = y , then g is not continuous but
g (V ) is open in X whenever V is open in Y .
Proof. Since every set is open in X , we have f 1 (V ) = g (V ) open for every
V in Y and so, in particular, for every open set. Thus f is continuous.
We observe that U = {0} is open in X and g 1 (U ) = f (U ) = U = {0} is
not open in Y . Thus g is not continuous.
[Return to page 10.]
Exercise 3.11. Consider R2 . For each of the British rail express and British
rail stopping metrics:(i) Describe the open balls. (Consider both large and small radii.)
(ii) Describe the open sets as well as you can. (There is a nice description
for the British rail express metric.) Give reasons for your answers.
51
Solution. We start with the British rail express metric. Write
BE ( ) = {x : x
2
< }
for the Euclidean ball centre 0 [ > 0]. If 0 < r < x 2 , then
B (x, r) = {x}.
If x
2
> r > 0, then
B (x, r) = {x} BE (r x ).
Since open balls are open and the union of open sets is open, we deduce that
every set not containing 0 and every set containing BE ( ) for some > 0 is
open.
On the other hand, if U is open and 0 U then U must contain BE ( )
for some > 0. It follows that the collection of sets described in the last
sentence of the previous paragraph constitute the open sets for the British
rail express metric.
We turn now to the stopping metric. We observe that
B (0, r) = BE (r)
for r > 0. If x = 0 and 0 < r < x 2 , then
B (x, r) =
If x = 0 and x
2
x
x
: ( x
2
r, x
2
+ r) .
2
> r > 0, then
B (x, r) =
x
x
: (0, x
2
+ r) BE (r x ).
2
A similar argument to the previous paragraph shows that the open sets are
precisely the unions of sets of the form
l(e, (a, b)) = {e : (a, b)}
where e is a unit vector and 0 a < b and unions sets of the form l(e, (a, b))
together with some BE ( ) with > 0.
[Return to page 10.]
Lemma 4.2. Consider a metric space (X, d). If a sequence xn has a limit
then that limit is unique.
52
Proof. Suppose xn x and xn y . Then, given any > 0, we can nd N1
and N2 such that
d(xn , x) < /2 for all n N1 and d(xn , y ) < /2 for all n N2 .
Taking N = max(N1 , N2 ), we obtain
d(x, y ) d(xN , x) + d(xN , y ) < /2 + /2 = .
Since was arbitrary, d(x, y ) = 0 and x = y .
[Return to page 11.]
Theorem 4.5. Let (X, d) be a metric space. A set F in X is closed if and
only if its complement is open.
Proof. Only if Suppose that F is closed and E = X \ F . If E is not open, we
can nd an e E such that B (e, ) F = for all > 0. In particular, we
can nd xn F such that d(xn , e) < 1/n for each n 1. Since xn e and
F is closed, we have e F contradicting our initial assumption that e E .
Thus E is open.
If We now establish the converse. Suppose E is open and F = X \ E .
Suppose xn F and xn x. If x E , then, since E is open we can nd a
> 0 such that B (x, ) E . Thus d(xn , x) for all n which is absurd.
Thus x F and F is closed.
[Return to page 11.]
Theorem 4.6. If (X, d) is a metric space, then the following statements are
true.
(i) The empty set and the space X are closed.
(ii) If F is closed for all A, then A F is closed. (In other words
the intersection of closed sets is closed.)
(iii) If Fj is closed for all 1 j n, then n=1 Fj is open.
j
Proof. (i) Observe that = X \ X and X = X \ .
(ii) Since F is closed, X \ F is open for all A. It follows that
X\
(X \ F )
F =
A
A
is open and so A F is closed.
(iii) Since Fj is closed, X \ Fj is open for all 1 j n. It follows that
n
X\
n
(X \ Fj )
Fj =
j =1
j =1
n
j =1
is open and so
Fj is closed.
[Return to page 11.]
53
Theorem 4.7. Let (X, d) and (Y, ) be metric spaces. A function f : X Y
is continuous if and only if f 1 (F ) is closed in X whenever F is closed in
Y.
Proof. If Suppose that f is continuous. If F is closed in Y , then Y \ F is
open, so
X \ f 1 (F ) = f 1 (Y \ F )
is open. Thus f 1 (F ) is closed.
Only if Suppose f 1 (F ) is closed whenever F is. If U is open in Y , then
Y \ U is closed, so
X \ f 1 (U ) = f 1 (Y \ U )
is closed. Thus f 1 (U ) is open. We have shown that f is continuous.
[Return to page 11.]
Exercise 5.6. Write P (Y ) for the collection of subsets of Y . If X has three
elements, how many elements does P P (X ) have?
How many topologies are there on X ?
3
Solution. If Y has n elements P (Y ) has 2n elements so P P (X ) has 22 =
28 = 256 elements.
Let X = {x, y, z }. We set out the types of possible topologies below.
type
{, X }
{, {x}, X }
{, {x}, {y }, {x, y }, X }
P (X )
{, {x, y }, X }
{, {x}, {x, y }, X }
{, {z }, {x, y }, X }
{, {x}, {z }, {x, y }, {x, z }, X }
{, {x}, {x, y }, {x, z }, X }
number of this type
1
3
3
1
3
6
3
6
3
There are that 29 distinct topologies on X .
The moral of this question is that although there are far fewer topologies than simple collections of subsets and even fewer dierent types (nonhomeomorphic topologies in later terminology) there are still quite a lot even
for spaces of three points.
[Return to page 13.]
54
Example 6.4. Let X = R and let d be the usual metric on R. Let Y = (0, 1)
(the open interval with end points 0 and 1) and let be the usual metric on
(0, 1). Then (X, d) and (Y, ) are homeomorphic as topological spaces but
(X, d) is complete and (Y, ) is not.
Proof. We know from rst year analysis that f (x) = tan( (y 1/2)) is a
bijective function f : Y X which is continuous with continuous inverse.
(Recall that a strictly increasing continuous function has continuous inverse.)
Thus (X, d) and (Y, ) are homeomorphic. We know that (X, d) is complete
by the general principle of analysis.
However 1/n is a Cauchy sequence in Y with no limit in Y . (If y (0, 1),
then there exists an N with y > N 1 . If m 2N , then |1/m y | 1/2N
so 1/n
y .)
[Return to page 14.]
Lemma 6.5. Let X be a space and let H be a collection of subsets of X .
Then there exists a unique topology H such that
(i) H H, and
(ii) if is a topology with H, then H .
Proof. The proof follows a standard pattern which is worth learning.
Uniqueness Suppose that and are topologies such that
(i) H,
(ii) if is a topology with H, then ,
(i) H,
(ii) if is a topology with H, then .
By (i) and (ii) , we have and by (i) and (ii), we have . Thus
= .
Existence Let T be the set of topologies with H. Since the discrete
topology contains H, T is non-empty. Set
H =
.
T
By construction H H and H whenever T . Thus we need only
show that H is a topology and this we now do.
(a) , X for all T so , X H .
(b) If U H then U for all A and so A U for all
T whence A U H .
(c) If Uj H then Uj for all 1 j n and so n=1 Uj for all
j
T whence n=1 Uj H .
j
Thus H is a topology, as required.
[Return to page 15.]
55
Lemma 6.8. If (X, ) is a topological space and Y X then the subspace
topology Y on Y is the collection of sets Y U with U .
Proof. Let j : Y X be the inclusion map given by j (y ) = y for all y Y .
Write
= {Y U : U }.
Since Y U = j 1 (U ) we know that Y is the smallest topology containing
and that the result will follow if we show that is a topology on Y . The
following observations show this and complete the proof.
(a) = Y and Y = Y X .
(b) A (Y U ) = Y A U .
(c) n=1 (Y Uj ) = Y n=1 Uj .
j
j
[Return to page 15.]
Lemma 6.12. Let (X, ) and (Y, ) be topological spaces and the product
topology on X Y . Then O if and only if, given (x, y ) O, we can nd
U and V such that
(x, y ) U V O.
Proof. Let be the collection of subsets E such that, given (x, y ) E , we
can nd U and V with
(x, y ) U V E.
If U then, since X is continuous U Y = X1 (U ) . Similarly, if
V then X V . Thus
U V = U Y X V .
If E then, given (x, y ) E , we can nd U(x,y) and V(x,y) such
that
(x, y ) U(x,y) V(x,y) E, .
We observe that
E
U(x,y) V(x,y) E
(x,y )E
so E = (x,y)E U(x,y) V(x,y) and, since the union of open sets is open, E .
Thus .
It is easy to check that is a topology as follows.
(a) vacuously. If (x, y ) X Y , then X , Y and
(x, y ) X Y X Y . Thus X Y .
56
(b) Suppose E for all A. If (x, y ) A E , then (x, y ) E
for some A. We can nd U and V such that
(x, y ) U V E
and so
(x, y ) U V
E .
A
Thus A E .
(c) Suppose Ej for all 1 j n. If (x, y ) n=1 Ej , then (x, y ) Ej
j
for all 1 j n. We can nd Uj and Vj such that
(x, y ) Uj Vj Ej
and so
n
(x, y )
n
Uj
j =1
n
Vj
j =1
Ej .
j =1
Since n=1 Uj and n=1 Vj , we have shown that
j
j
Finally, we observe that, if U , then
n
j =1
Ej .
X1 (U ) = U Y
and (x, y ) U Y X1 (U ) with U , Y , so X1 (U ) . Thus
X : X Y X is continuous if we give X Y the topology . A similar
result holds for Y so, by the minimality of , = .
[Return to page 16.]
Lemma 6.13. Let 1 and 2 be two topologies on the same space X .
(i) We have 1 2 if and only if, given x U 1 , we can nd V 2
such that x V U .
(ii)We have 1 = 2 if and only if, given x U 1 , we can nd V 2
such that x V U and, given x U 2 , we can nd V 1 such that
x V U.
Proof. (i) If 1 2 and x U 1 , then setting V = U we automatically
have V 2 and x V U .
Conversely, suppose that, given x U 1 , we can nd V 2 such
that x V U . Then, if U 1 is xed, we can nd Vx 2 such that
x Vx U for each x U .
Now
U
Vx U
x U
57
so U = xU Vx and, since the union of open sets is open, U 2 . Thus
1 2 .
(ii) Observe that 1 = 2 if and only if 1 2 and 2 1 .
[Return to page 16.]
Exercise 7.4. If (X, ) is a topological space, then a subset A of X is open
if and only if every point of A has an open neighbourhood U A.
Proof. If A is open, then A is an open neighbourhood of every x A.
Conversely, suppose that every x A has an open neighbourhood Ux
lying entirely within A. Then
A
Ux A
x A
so A = xA Ux . Thus A is the union of open sets and so open.
[Return to page 18.]
Lemma 7.5. If (X, ) is a Hausdor space, then the one point sets {x} are
closed.
Proof. We must show that A = X \ {x} is open. But, if y A then y = x so
by, the Hausdor condition, we can nd U, V such that x U , y V
and U V = . We see that y V A, so every point of A has an open
neighbourhood lying entirely within A. Thus A is open.
[Return to page 18.]
Exercise 7.6. Let X be innite (we could take X = Z or X = R). We say
that a subset E of X lies in if either E = or X \ E is nite. Show that
is a topology and that every one point set {x} is closed but that (X, ) is
not Hausdor.
What happens if X is nite?
Proof. (a) We are told that . Since X \ X = , X .
(b) If U for all A then, either U = for all A so
A U = or we can nd a A such that X \ U is nite. In the
second case we observe that
X\
U X \ U ,
A
so X \
A
U is nite and
A
U
58
(c) If Uj for all 1 j n then, either Uk = for some 1 k n
so n=1 Uj = or X \ Uj is nite for all 1 j n. In the second case
j
then, since
n
X\
n
(X \ Uj ),
Uj =
j =1
j =1
it follows that X \ n=1 Uj is nite and so n=1 Uj .
j
j
Thus is a topology.
Since {x} is nite X \ {x} is open and so {x} is closed.
Suppose that x = y and x U , y V . Then U, V = so X \ U
and X \ V is nite. It follows that
X \ U V = (X \ U ) (X \ V )
is nite, and so, since X is innite, U V = . Thus is not Hausdor.
If X is nite then is the discrete metric which is Hausdor.
[Return to page 19.]
Lemma 7.7. If (X, ) is a Hausdor topological space and Y X , then Y
with the subspace topology is also Hausdor.
Proof. Write Y for the subspace topology. If x, y Y and x = y , then
x, y X and x = y so we can nd U, V with x U , y V and
U V = . Set U = U Y and V = V Y . Then U , V Y x U , y V
V = .
and U
[Return to page 19.]
Lemma 7.8. If (X, ) and (Y, ) are Hausdor topological spaces, then X Y
with the product topology is also Hausdor.
Proof. Suppose (x1 , y1 ), (x2 , y2 ) and (x1 , y1 ) = (x2 , y2 ). Then we know that
at least one of the statements x1 = x2 and y1 = y2 is true9 . Without loss of
generality we may suppose x1 = x2 . Since (X, ) is Hausdor we can nd
U1 , U2 disjoint open neighbourhoods of x1 and x2 . We observe that U1 Y
and U2 Y are disjoint open neighbourhoods of (x1 , y1 ) and (x2 , y2 ) so we
are done.
[Return to page 19.]
Exercise 8.4. (iv) Show that the topology described in Exercise 7.6 is compact.
(v) Let X be uncountable (we could take X = R). We say that a subset
A of X lies in if either A = or X \ A is countable. Show that is a
topology but that (X, ) is not compact.
9
But not necessarily both. This is the traditional silly mistake.
59
Solution. (iv) If X = there is nothing to prove. If not, let U [ A]
be an open cover. Since X = we can choose a A such that U =
and so U = X \ F where F is a nite set. For each x F we know that
x X = A U so there exists an (x) A with x U(x) . We have
U
U(x) = X,
x F
giving us the desired open cover.
(v) I leave it the reader to show that is a topology. Let x1 , x2 , . . . , be
distinct points of X . Let
U = X \ {xj : 1 j }
and Uk = U {xk }. Then Uk [k 1] and k1 Uk = X . Now suppose
k (1), k (2),. . . , k (N ) given. If m = max1rN k (r), then
N
xm+1
/
Uk(r)
r =1
so there is no nite subcover.
[Return to page 20.]
Theorem 8.5. [The HeineBorel Theorem.] Let R be given its usual
(Euclidean) topology. Then the closed bounded interval [a, b] is compact.
Proof. Suppose that C is an open cover of [a, b] (i.e. the elements of C are
open sets and U C U [a, b]). If C1 is a nite subcover of [a, c] and C2 is a
nite subcover of [c, b] then C1 C2 is a nite subcover of [a, b].
Suppose now that [a, b] has no nite subcover using C . Set a0 = a, b0 = b,
and c0 = (a0 + b0 )/2. By the rst paragraph at least one of [a0 , c0 ] and [c0 , b0 ]
has no nite subcover using C . If [a0 , c0 ] has no nite subcover, set a1 = a0 ,
b1 = c0 . Otherwise, set a1 = c0 , b1 = b0 . In either case, we know that
(i) a = a0 a1 b1 b0 = b,
(ii) if F is a nite subset of C , then U F U [a1 , b1 ],
(iii) b1 a1 = (b a)/2.
Proceeding inductively, we obtain
(i)n a an1 an bn bn1 b.
(ii)n If F is a nite subset of C , then U F U [an , bn ].
(iii)n bn an = 2n (b a).
The an form an increasing sequence bounded above by b, so, by the fundamental axiom of analysis, an A for some A b. Similarly bn B for
60
some B a. Since bn an 0, A = B = x, say, for some x [a, b]. Since
x [a, b] and U C U [a, b] we can nd a V C with x V . Since V is
open in the Euclidean metric, we can nd a > 0 such that (x , x + ) V .
Since an , bn x we can nd an N such that |x aN |, |x bN | < and so
[aN , bN ] (x , x + ) V
contradicting (ii)N . (Just take F = {V }.)
The theorem follows by reductio ad absurdum.
[Return to page 20.]
Theorem 8.7. A closed subset of a compact set is compact. [More precisely,
if E is compact and F closed in a given topology, then, if F E , it follows
that F is compact.]
Proof. Suppose (X, ) is a topological space, E is a compact set in X and F
is a closed subset of E . If U [ A] and A U F , then X \ F
and
(X \ F )
U = X E.
A
By compactness, we can nd (j ) A [1 j n] such that
n
(X \ F )
U(j ) E.
j =1
Since (X \ F ) F = and E F , it follows that
n
U(j ) F
j =1
and we are done. [Return to page 21.]
Theorem8.8. If (X, ) is Hausdor, then every compact set is closed.
Proof. Let K be a compact set. If x K , then, given, any k K we know
/
that k = x and so, since X is Hausdor, we can nd open sets Uk and Vk
such that
x Vk , k Uk and Vk Uk = .
Since kK Uk kK {k } = K , we have an open cover of K . By compactness, we can nd k (1), k (2), . . . , k (n) K such that
n
Uk(j ) K.
j =1
61
n
j =1
We observe that the nite intersection V =
hood of x and that
n
V K V
Vk(j ) is an open neighbour-
Uk(j ) = ,
j =1
so V K and we have shown that every x X \ K has an open neighbourhood
lying entirely within X \ K . Thus X \ K is open and K is closed.
[Return to page 21.]
Example 8.9. Give an example of a topological space and a compact set
which is not closed.
Proof. If (X, ) has the indiscrete topology, then, if Y X , Y = X, , we
have Y compact but not closed. We can take X = {a, b} with a = b and
Y = {a}.
[Return to page 21.]
Theorem 8.10. Consider (R, ) with the standard (Euclidean) topology. A
set E is compact if and only if it is closed and bounded (that is to say, there
exists a M such that |x| M for all x E ).
Proof. If E is bounded, then E [M, M ] for some M . By the theorem of
HeineBorel, [M, M ] is compact so, if E is closed, E is compact.
Since (R, ) is Hausdor any compact set must be closed. Finally suppose
that E is compact. We have
E
(j, j ).
j =1
By compactness, we can nd j (r) such that E N (j (r), j (r)) Taking
r =1
M = max1rn j (r) we have E (M, M ) so E is bounded.
[Return to page 21.]
Theorem 8.12. Let (X, ) and (Y, ) be topological spaces and f : X Y a
continuous function. If K is a compact subset of X , then f (K ) is a compact
subset of Y .
Proof. Suppose that U for all A and
f 1 (U ) = f 1
A
U
A
62
A
U f (K ). Then
K
and, since f is continuous f 1 (U ) for all A. By compactness, we
can nd (j ) A [1 j n] such that
n
f 1 (U(j ) ) K
j =1
and so
n
n
f 1 (U(j ) )
U(j ) = f
j =1
f (K )
j =1
and we are done.
[Return to page 22.]
Theorem 8.15. Let R have the usual metric. If K is a closed and bounded
subset of R and f : K R is continuous, then f is bounded and attains its
bounds.
Proof. If K is empty there is nothing to prove, so we assume K = .
Since K is compact and f is continuous f (K ) is compact. Thus f (K )
is a non-empty closed bounded set. Since f (K ) is non-empty and bounded,
it has a supremum , say. Since f (K ) is closed, it contains its supremum.
[Observe that we can nd kn K such that
1/n f (kn )
and so f (kn ) . Since f (K ) is closed, f (K ).]
[Return to page 22.]
Exercise 8.16. Let R have the usual metric. If K is a subset of R with the
property that, whenever f : K R is continuous, then f is bounded and
attains its bounds, it follows that K is closed and bounded.
Proof. If K = there is nothing to prove, so we assume K = .
Let f : K R be dened by f (k ) = |k |. Since f is bounded, K must be.
If x K , then the function f : K R given by f (k ) = |k x|1 is
/
continuous and so bounded. Thus we can nd an M > 0 such that |f (k )| <
M for all k K . It follows that |x k | > M 1 for all k K and the open
ball B (x, M 1 ) lies entirely in the complement of K . Thus K is closed.
[Return to page 22.]
Theorem 8.18. Let (X, ) be a compact and (Y, ) a Hausdor topological
space. If f : X Y is a continuous bijection, then it is a homeomorphism.
63
Proof. Since f is a bijection, g = f 1 is a well dened function. If K is
closed in X , then (since a closed subset of a compact space is compact) K is
compact so f (K ) is compact. But a compact subset of a Hausdor space is
closed so g 1 (K ) = f (K ) is closed. Thus g is continuous and we are done.
(If U is open in X then X \ U is closed so Y \ g 1 (U ) = g 1 (X \ U ) is closed
and g 1 (U ) is open.)
[Return to page 23.]
Theorem 8.20. Let 1 and 2 be topologies on the same space X .
(i) If 1 2 and 1 is compact, then so is 2 .
(ii) If 1 2 and 2 is Hausdor, then so is 1 .
(iii) If 1 2 , 1 is compact and 2 is Hausdor, then 1 = 2 .
Proof. (i) The map : (X, 1 ) (X, 2 ) is continuous and so takes compact
sets to compact sets. In particular, since X is compact, in 1 , X = X is
compact in 2 .
(ii) If x = y we can nd x U 2 and y V 2 with U V = .
Automatically x U 1 and y V 1 so we are done.
(iii) The map : (X, 1 ) (X, 2 ) is a continuous bijection and so a
homeomorphism.
[Return to page 23.]
Exercise 8.21. (i) Give an example of a compact space (X, ) and a topological space (Y, ) together with a continuous bijection f : X Y which is
not a homeomorphism.
(ii) Give an example of a topological space (X, ) and a Hausdor space
(Y, ) together with a continuous bijection f : X Y which is not a homeomorphism.
Solution. Let 1 be the indiscrete topology on [0, 1], 2 be the usual (Euclidean) topology on [0, 1] and 3 be the discrete topology on [0, 1]. Then
(R, 1 ) is compact (but not Hausdor), (R, 2 ) is compact and Hausdor,
and (R, 3 ) is Hausdor (but not compact). The identity maps : (X, 1 )
(X, 2 ) and : (X, 2 ) (X, 3 ) are continuous bijections but not homeomorphisms.
[Return to page 23.]
Theorem 9.1. The product of two compact spaces is compact. (More formally, if (X, ) and (Y, ) are compact topological spaces and is the product
topology then (X Y, ) is compact.)
64
Proof. Let O [ A] and
O = X Y.
A
Then, given (x, y ) X Y , we can nd Ux,y , Vx,y and (x, y ) A
such that
(x, y ) Ux,y Vx,y O(x,y) .
In particular, we have
{x} Vx,y = {(x, y ) : y Y }
y Y
for each x X and so
Vx,y = Y.
y Y
By compactness, we can nd a positive integer n(x) and y (x, j ) Y [1
j n(x)] such that
n(x)
Vx,y(x,j ) = Y.
j =1
(x
Now Ux = n=1) Ux,y(x,j ) is the nite intersection of open sets in X and so
j
open. Further x Ux and so
Ux = X.
x X
By compactness, we can nd x1 , x2 , . . . , xm such that
m
Uxr = X.
r =1
It follows that
m n(xr )
m n(xr )
Oxr ,y(xr ,j )
r =1 j =1
Uxr ,y(xr ,j ) Vxr ,y(xr ,j )
r =1 j =1
m n(xr )
Uxr Vxr ,y(xr ,j )
r =1 j =1
m
Uxr Y
r =1
X Y
65
and we are done.
[Return to page 24.]
Exercise 9.7. Consider the complex plane with its usual metric. Let
D = {z C : |z | = 1}
and give D the subspace topology . Give R its usual topology and dene
an equivalence relation by x y if x y Z. We write R/ = T and give
T the quotient topology. The object of this exercise is to show that D and
T are homeomorphic.
(i) Verify that is indeed an equivalence relation.
(ii) Show that, if we dene f : R D by f (x) = exp(2ix), then f (U )
is open whenever U is open.
(iii) If q : R T is the quotient map q (x) = [x] show that q (x) = q (y )
if and only if f (x) = f (y ). Deduce that q f 1 ({exp(2ix)}) = [x] and that
the equation F (exp(2ix)) = [x] gives a well dened bijection F : D T.
(iv) Show that F 1 (V ) = f q 1 (V ) and deduce that F is continuous.
(v) Show that T is Hausdor and explain why D is compact. Deduce
that F is a homeomorphism.
Solution. (i) Observe that x x = 0 Z so x x.
Observe that x y implies x y Z so y x = (x y ) Z and y x.
Observe that, if x y and y z , then x y, y z Z so
x z = (x y ) + (y z ) = x z Z
and x z .
(ii) If x U an open set, then we can nd a 1 > > 0 such that |x y | <
implies y U .
By simple geometry, any z C with |z | = 1 and | exp(2ix) z | < /100
can be written as z = exp(2iy ) with |y x| < . Thus
D {z C : |z exp(2ix)| < /100} f (U ).
We have shown that f (U ) is open.
(iii) We have
q (x) = q (y ) y [x] x y Z exp(2i(x y )) = 1
exp(2ix) = exp(2iy ) f (x) = f (y ).
It follows that the equation F (exp(2ix)) = [x] gives a well dened bijection F : D T.
66
(iv) Observe that
F 1 ([x]) = {exp(2it) : exp(2it) = exp(2ix)} = f q 1 ([x])
and so F 1 (V ) = f q 1 (V ) . If V is open, then, since q is continuous, q 1 (V )
is open so, by (ii), F 1 (V ) is open. Thus F is continuous.
(v) If [x] = [y ], then we know that x y Z and the set
/
{|t| : t (x y ) Z, |t| < 1}
is nite and non-empty. Thus there exists a > 0 such that
{|t| : t (x y ) Z, |t| < } = {}.
Let
Ux =
(j + x /4, j + x + /4) and Uy =
j =
(j + y /4, j + y + /4).
j =
Observe that Ux and Uy are open in R and q 1 q (Ux )) = Ux , q 1 q (Uy )) = Uy ,
and so q (Ux ) and q (Uy ) are open in the quotient topology. Since [x] q (Ux ),
[y ] q (Uy ) and q (Ux ) q (Uy ) = , we have shown that the quotient topology
is Hausdor.
Since D is closed and bounded in C and we can identify C with R2 as a
metric space, D is compact.
Since a continuous bijection from a compact to a Hausdor space is a
homeomorphism, F is a homeomorphism.
[Remark. It is just as simple to show that the natural map from T (which
we know to be compact, why?) to D (which we know to be Hausdor, why?)
is a bijective continuous map. Or we could show continuity in both directions
and not use the result on continuous bijections.]
[Return to page 25.]
Theorem 10.4. If E is a subset of a topological space (X, ), then E is
disconnected if and only if we can nd a non-constant continuous function
f : E R which only takes the value 0 or 1.
Proof. Since we are dealing with a subspace topology, we can take E = X .
If f : X R is a continuous non-constant function which only takes the
value 0 or 1, then U = f 1 ({0}) = f 1 ((1/2, 1/2)) is open and non-empty
and similarly V = f 1 ({1}) is. Since V U = X and V U = , it follows
that X is disconnected.
67
Conversely, if X is disconnected we can nd non-empty open sets U and
V such that V U = X and V U = . If we set f (u) = 0 when u U
and f (v ) = 1 when v V , then f : X R is a well dened non-constant
function which only takes the value 0 or 1. If A R, the f 1 (A) must be one
of the four sets , U , V or X all of which are open. Thus f is continuous.
[Return to page 26.]
Theorem 10.5. If we give R the usual topology, then R and the intervals
[a, b] and (a, b) are connected.
Proof. We prove the result for (a, b). The other results are proved similarly.
Suppose f : (a, b) R is continuous and there exist c, d (a, b) with
f (c) = 0 and f (d) = 1. Without loss of generality we may suppose that
c < d and so a < c < d < b. By the intermediate value theorem, we can nd
(c, d) with f ( ) = 1/2. Since (a, b), f takes at least three values and
(a, b) must be connected.
[Return to page 26.]
Exercise 10.6. (i) If (X, ) and (Y, ) are topological spaces, E is a connected subset of X and g : E Y is continuous, then g (E ) is connected.
(More briey the continuous image of a connected set is connected.)
(ii) If (X, ) is a connected topological space and is an equivalence
relation on X , then X/ with the quotient topology is connected.
(iii) If (X, ) and (Y, ) are connected topological spaces, then X Y with
the product topology is connected.
(iv) If (X, ) is a connected topological space and E is a subset of X , it
does not follow that E with the subspace topology is connected.
Proof. (i) If g (E ) is not connected we can nd a non-constant continuous f :
g (E ) R taking only the values 0 and 1. Setting F = f g (the composition
of f and g ), we know that F : E R is non-constant, continuous and only
takes the values 0 and 1. Thus E is not connected.
(ii) X/ is the continuous image of X under the quotient map which we
know to be continuous.
(iii) Suppose X Y with the product topology is not connected. Then
we can nd a non-constant continuous function f : X Y R taking only
the values 0 and 1. Take (x, y ), (u, v ) X Y with f (x, y ) = f (u, v ).
Then, if f (x, v ) = f (x, y ), it follows that f (x, v ) = f (u, v ). Without loss of
generality, suppose that f (x, v ) = f (x, y ). Then we know that the function
: Y X Y given by (z ) = (x, z ) is continuous. (If is open in X Y
and z 1 (), then (x, z ) so we can nd U open in X and V open
in Y such that (x, z ) U V . Thus z V 1 () and we have
68
shown 1 () open.) If we set F = f , then F : Y R is non-constant,
continuous and only takes the values 0 and 1. Thus Y is not connected.
(iv) R is connected with the usual topology but E = (2, 1) (1, 2) is
not.
[Return to page 26.]
Example 10.7. Suppose that E is a connected subset of a topological space
(X, ). Suppose that f : E R is locally constant in the sense that, given
any e E , we can nd an open neighbourhood U of e such that f is constant
on U E . Then f is constant.
Proof. Since we are dealing with the subspace topology on E , there is no loss
in generality in taking E = X . If X = the result is vacuous so we may
take X = .
Choose an x0 X and set c = f (x0 ). Now consider the sets
U = {x X : f (x) = c} and V = {x X : f (x) = c}.
We claim that U and V are open. For suppose v V . Then we can nd
an open neighbourhood N of v such that f is constant on N . Thus f (x) =
f (v ) = c for all x N , so N V . It follows that V is open. A similar,
slightly simpler, argument shows that U is open.
Since U V = , U V = X and U = the connectedness of X tells
us that V = and U = X . The result follows.
[Return to page 27.]
Lemma 10.9. We work in a topological space (X, ).
(i) Let x0 X . If x0 E and E is connected for all A, then
A E is connected.
(ii) Write x y if there exists a connected set E with x, y E . Then
is an equivalence relation.
(iii) The equivalence classes [x] are connected.
(iv) If F is connected and F [x], then F = [x].
Proof. (i) Let U and V be open sets such that
U V
E and U V
A
E = .
A
Without loss of generality, let x0 U . Then
U V E and U V E =
69
for each A. But x0 U E so U E = and so, by the connectedness
of E , we have
U E
for all A. Thus U A E . We have shown that A E is connected.
(ii) Observe that if U and V are sets (open or not) such that
U V {x}, and U V {x} = .
then either x U and U {x} = or x U so U {x}. Thus the one
/
point set {x} is connected and x x.
The symmetry of the denition tells us that, if x y , then y x.
If x y and y z , then x, y E and y, z F for some connected sets
E and F . By part (i), E F is connected (observe that y E, F ) so, since
x, z E F , x z .
We have shown that is an equivalence relation.
(iii) If y [x], then there exists a connected set Ey with x, y Ey . By
denition [x] Ey so
{y }
[x] =
y [x]
Ey [x]
y [x]
whence
[x] =
Ey
y [x]
and, by part (i), [x] is connected.
(iv) If F is connected and [x] F then x F and, by denition of ,
[x] F . It follows that F = [x].
[Return to page 27.]
Lemma 10.11. If (X, ) is a topological space and we write x y if x is
path-connected to y , then is an equivalence relation.
Proof. If x X , then the map : [0, 1] X dened by (t) = x for all t is
continuous. (Observe that, if F is a closed set in X, then 1 (F ) takes the
value or [0, 1] both of which are closed.) Thus x x.
If x y , then we can nd a continuous map : [0, 1] X with (0) = x
and (1) = y . The map T : [0, 1] [0, 1] given by T (t) = 1 t is continuous
so the composition = T is. Observe that (0) = y and (1) = x so
y x.
70
If x y and y z , then we can nd continuous maps j : [0, 1] X
with 1 (0) = x 1 (1) = y , 2 (0) = y and 2 (1) = z . Dene : [0, 1] X by
(t) =
1 (2t)
if t [0, 1/2]
2 (2t 1) if t (1/2, 1].
If U is open in X then
1 (U ) = {t/2 : t 1 1 (U )} {(1 + t)/2 : t 2 1 (U )}
is open.
(If more detail is required we argue as follows. Suppose s 1 (U ). If
s (0, 1/2), then 2s 1 1 (U ) so, since 1 1 (U ) is open we can nd a > 0
with s > such that (2s , 2s + ) 1 1 (U ). Thus (s /2, s + /2)
1 (U ). If s = 0 then 0 1 1 (U ) so, since 1 1 (U ) is open we can nd a > 0
1
with 1 > such that [0, ) 1 (U ). Thus [s, /2) = [0, /2) 1 (U ).
The cases s (1/2, 1] are dealt with similarly. This leaves the case s = 1/2.
Arguing as before, we can nd 1 , 2 > 0 with 1 > 1 , 2 such that
(1 1 , 1] 1 1 (U ) and [0, 2 ) 2 1 (U ).
Setting = min(1 , 2 ) we have
(s /2, s + /2) = (1/2 /2, 1/2 + /2) 1 (U ).
We see that the case s = 1/2 is really the only one which requires care.)
Thus is continuous and, since (0) = x, (1) = z , x z .
[Return to page 70.]
Theorem 10.12. If a topological space is path-connected, then it is connected.
Proof. Suppose that (X, ) is path-connected and that U and V are open
sets with U V = and U V = X . If U = , choose x U . If y X ,
we can nd f : [0, 1] X continuous with f (0) = x and f (1) = y . Now the
continuous image of a connected set is connected and [0, 1] is connected, so
f ([0, 1]) is connected. Since
U V f ([0, 1]) = , U V f ([0, 1]) and U f ([0, 1]) = ,
we know that U f ([0, 1]) so y U . Thus U = X . We have shown that X
is connected.
[Return to page 28.]
71
Theorem 10.13. If we give Rn the usual topology then any open set which
is connected is path-connected.
Proof. If = there is nothing to prove, so we assume non-empty.
Pick x and let U be the set of all points in which are path-connected
to x and let V be the set of all points in which are not. We shall prove
that U and V are open.
Suppose rst that u U . Since is open, we can nd an open ball
B (u, ) centre u, radius > 0 lying entirely within . If y B (u, ), then
u is path-connected to y in B (u, ) and so in U . (Consider : [0, 1]
given by (t) = tu + (1 t)y.) Since x is path-connected to u and u is
path-connected to y, it follows that x is path-connected to y in so y U .
Now suppose that v V . Since is open, we can nd an open ball
B (v, ) centre v, radius > 0 lying entirely within . If y B (v, ), then
v is path-connected to y in B (v, ) and so in V . It follows that, if y is
path-connected to x, then so is v. But v V , so y is not path-connected to
x. Thus y V .
Since U V = and U V = , the connectedness of shows that
U = and is path-connected.
[Return to page 28.]
Example 10.14. We work in R2 with the usual topology. Let
E1 = {(0, y ) : |y | 1} and E2 = {(x, sin 1/x) : 0 < x 1}
and set E = E1 E2 .
(i) Sketch E .
(ii) Explain why E1 and E2 are path-connected and show that E is closed
and connected.
(iii) Suppose, if possible, that x : [0, 1] E is continuous and x(0) =
(1, 0), x(1) = (0, 0). Explain why we can nd 0 < t1 < t2 < t3 < . . . such
1
that x(tj ) = (j + 2 ) )1 . By considering the behaviour of tj and y (tj ), obtain
a contradiction.
(iv) Deduce that E is not path-connected.
Proof. Part (i) is left to the reader.
(ii) If y1 , y2 [1, 1], the function f : [0, 1] E1 given by
f (t) = 0, (1 t)y1 + ty2
is continuous and f (0) = (x1 , 0) and f (1) = (x2 , 0), so E1 is path-connected.
72
If (x1 , y1 ), (x2 , y2 ) E2 , then yj = sin 1/xj and setting
g(t) = (1 t)x1 + tx2 , sin 1/((1 t)x1 + tx2 )
we see that g is continuous and g(0) = (x1 , y1 ) and g(1) = (x2 , y2 ), so E2 is
path-connected.
We next show that E is closed. Suppose that (xr , yr ) E and (xr , yr )
(x, y ). If x = 0, then we note that, since |yr | 1 for all r and yr y , we
have |y | 1 and (x, y ) E1 E . If x = 0, then 1 x > 0 (since xr 0 for
all r). We can nd an N such that |x xr | < x/2 and so xr > x/2 for all
r N . Thus, by continuity,
(xr , yr ) = (xr , sin 1/xr ) (x, sin 1/x) E2 E.
Thus E is closed.
Now suppose, if possible, that E is disconnected. Then we can nd U
and V open such that
U E = , V E = , U V E and U V E = .
Then
U V Ej and U V Ej = .
and so, since Ej is path-connected, so connected, we have U Ej = or
V Ej = [j = 1, 2]. Without loss of generality, assume V E1 = so
U E1 . Since (0, 0) E1 , we have (0, 0) U . Since U is open, we can nd
a > 0 such that (x, y ) U whenever (x, y ) 2 < . If n is large,
((n )1 , 0) U E2 = U V E,
contradicting our initial assumptions. By reductio ad absurdum, E is connected.
(iii) Write x(t) = (x(t), y (t)). Since x is continuous so is x. Since x(0) = 1
and x(1) = 0, the intermediate value theorem tells us that we can nd t1 with
0 < t1 < 1 and x(t1 ) = ( 3 )1 . Applying the intermediate value theorem
2
again, we can nd t2 with 0 < t2 < t1 and x(t2 ) = ( 5 )1 . We continue
2
inductively.
Since the tj form a decreasing sequence bounded below by 0, we have
tj T for some T [0, 1]. Since y is continuous
(1)j = sin 1/x(tj ) = y (tj ) y (T )
which is absurd.
(iv) Part (iii) tells us that there is no path joining (0, 0) and (1, 0) in E ,
so E is not path-connected.
[Return to page 28.]
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Example 11.1. Give an example of metric space (X, d) which is bounded
(in the sense that there exists an M with d(x, y ) M for all x, y X ) but
for which there exist sequences with no convergent subsequence.
Solution. Consider the discrete metric on Z. If xn = n and x Z, then
d(x, xn ) = 1 for all n with at most one exception. Thus the sequence xn can
have no convergent subsequence.
[Return to page 29.]
Theorem 11.4. If the metric space (X, d) is compact, it is sequentially compact.
Proof. Let xn be a sequence in X . If it has no convergent subsequence, then,
for each x X we can nd a (x) > 0 and an N (x) such that xn B (x, (x))
/
for all n N (x). Since
{x}
X=
x X
B (x, (x)) X,
x X
the B (x, (x)) form an open cover and, by compactness, have a nite subcover. In other words, we can nd an M and yj X [1 j M ] such
that
M
X=
B yj , (yj ) .
j =1
Now set N = max1j M N (yj ). Since N N (yj ), we have xN
/
M
B yj , (yj ) for all 1 j M . Thus xN j =1 B yj , (yj ) = X which is
/
absurd.
The result follows by reductio ad absurdum.
[Return to page 29.]
Lemma 11.5. Suppose that (X, d) is a sequentially compact metric space
and that the collection U with A is an open cover of X . Then there
exists a > 0 such that, given any x X , there exists an (x) A such
that the open ball B (x, ) U(x) .
Proof. Suppose the rst sentence is true and the second sentence false. Then,
for each n 1 we can nd an xn such that the open ball B (xn , 1/n) U
for all A. By sequential compactness, we can nd y X and n(j )
such that xn(j ) y .
Since y X , we must have y U for some A. Since U is open,
we can nd an such that B (y, ) U . Now choose J suciently large
74
that n(J ) > 21 and d(xn(J ) , y ) < /2. We now have, using the triangle
inequality, that
B (xn(J ) , 1/n(J )) B (xn(J ) , /2) B (y, ) U ,
contradicting the denition of xn(J ) .
The result follows by reductio ad absurdum.
[Return to page 29.]
Theorem 11.6. If the metric space (X, d) is sequentially compact, it is compact.
Proof. Let (U )A be an open cover and let be dened as in Lemma 11.5.
The B (x, ) form a cover of X . If they have no nite subcover, then given x1 ,
x2 , . . . xn we can nd an xn+1 n=1 B (xj , ). Consider the sequence xj thus
/j
obtained. We have d(xn+1 , xk ) > whenever n k 1 and so d(xr , xs ) >
for all r = s. It follows that, if x X , d(xn , x) > /2 for all n with at most
one exception. Thus the sequence of xn has no convergent subsequence.
It thus follows, by reductio ad absurdum, that the B (x, ) have a nite
subcover. In other words, we can nd an M and yj X [1 j M ] such
that
M
X=
B (yj , ).
j =1
We thus have
M
M
B (yj , )
X=
j =1
U(yj ) X
j =1
so X = M U(yj ) and we have found a nite subcover.
j =1
Thus X is compact.
[Return to page 29.]
Lemma 12.2. Let (X, ) be a topological space. Then U if and only if,
given x U , we can nd a neighbourhood N of x with N U .
Proof. If U then U is a neighbourhood of x for all x U .
Conversely, if given any x U , we can nd a neighbourhood Nx of x with
Nx U , then we can nd an open neighbourhood Ux of x with Ux Nx .
Since
U
{x}
Ux
Nx
U = U,
x U
x U
x U
we have U = xU Ux .
[Return to page 31.]
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x U
Lemma 12.3. Let (X, ) and (Y, ) be topological spaces. Then f : X Y
is continuous if and only if, given x X and M a neighbourhood of f (x) in
Y , we can nd a neighbourhood N of x with f (N ) M .
Proof. If If f : X Y is continuous, x X and M is a neighbourhood of
f (x), then we can nd a V with f (x) V M . Since f is continuous
f 1 (V ) . Thus, since x f 1 (V ), we have that f 1 (V ) is an open
neighbourhood and so a neighbourhood of x. Setting N = f 1 (V ), we have
f (N ) = V M as required.
Only if Suppose that, given x X and M a neighbourhood of f (x) in Y ,
we can nd a neighbourhood N of x with f (N ) M . Let V be open in
Y . If x X and f (x) V , then V is a neighbourhood of f (x) so there
exists a neighbourhood Nx of x with f (Nx ) V . We now choose Ux an open
neighbourhood of x with Ux Nx . We have
f (Ux ) V
and so Ux f 1 (V ) for all x f 1 (V ). Thus
f 1 (V ) =
{x}
x f 1 ( V )
It follows that f 1 (V ) =
uous.
[Return to page 31.]
17
f 1 (V ) = f 1 (V ).
Ux
x f 1 ( V )
x f 1 ( V )
x f 1 ( V )
Ux . We have shown that f is contin-
Executive summary
Metrics
Denition and examples [Page 3]. Continuity [Page 7]. Open sets [Page 8].
Characterising continuous functions using open sets [Theorem 3.7, Page 9].
Limits [Page 10]. Closed sets [Page 11].
Topology
Denition of a topology [Page 12]. Metric topologies [Theorem 5.2, Page 12].
Further examples [Page 12]. Continuous functions [Page 13] and closed sets
[Page 13]. Homeomorphisms [Page 14]. Topological and non-topological
properties [Page 14] illustrated by completeness [Page 14]. Subspace [Page 15]
product [Page 16] and quotient [Lemma 6.15, Page 17] topologies. Hausdor
spaces [Page 18].
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Compactness
Denition using open sets [Page 20]. Examples: nite sets [Example 8.4,
Page 20] and [0, 1] [Theorem 8.5, Page 20]. Closed subsets of compact sets
are compact [Theorem 8.7, Page 21]. Compact subsets of a Hausdor space
must be closed [Theorem 8.8, Page 21]. The compact subsets of the real line
[Theorem 8.10, Page 21]. Continuous images of compact sets are compact
[Theorem 8.12, Page 22]. Quotient spaces [Page 22]. Continuous real valued
functions on a compact space are bounded and attain their bounds [Theorem 8.17, Page 22]. The product of two compact spaces is compact [Theorem 9.1, Page 24]. The compact subsets of Euclidean space [Theorem 9.5,
Page 25]. Sequential compactness [Page 29].
Connectedness
Denition using open sets [Page 26] and integer valued functions [Theorem 10.4, Page 26]. Examples, including intervals [Theorem 10.5, Page 26].
Continuous image of a connected set is connected [Example 10.6 (i), Page 26].
Components [Lemma 10.9, Page 27]. Path-connectedness [Page 27]. Pathconnected spaces are connected [Theorem 10.12, Page 28] but not conversely
[Example 10.14, Page 28]. Connected open sets in Euclidean space are pathconnected [Theorem 10.13, Page 28].
Neighbourhoods
Open neighbourhoods [Page 18]. Neighbourhoods [Page 31]. Continuity
via neighbourhoods [Lemma 12.3, Page 31]. Neighbourhoods [Exercise 12.4,
Page 31] and convergence [Lemma 12.5, Page 31] in metric spaces. Limits
treacherous concept in general topological spaces [Page 31].
77